Syllbus Acids nd Bses, ph, Common ion effect, Buffer solutions, Hydrolysis of slts nd Solubility Product. Acids nd Bses Here we discuss some importnt definitions of cids nd bses. Arrhenius Definition Arrhenius defined n cid to be substnce which when dissolved in wter, dissocites to give H + ions. A bse ws defined s substnce which when dissolved in wter, dissocites to give OH ions. Bronsted Lowry Definition According to this definition n cid is proton donor nd bse is proton cceptor. This definition mkes no mention of the solvent (nd is pplicble even if no solvent is present). ph of Acids nd Bses Mny properties of queous solutions depends on the concentrtion of H + ions of the solutions nd therefore there is need to express these concentrtions in simple terms. For this purpose we introduce the concept of ph. ph = log H + (Where H + is the ctivity of H + ions). Activity of H + ions is the concentrtion of free H + ions in solution. By free, we men those tht re t lrge distnce from the other ion so s not to experience its pull. We cn infer from this tht in dilute solutions, the ctivity of n ion is sme s its concentrtion since more number of solvent molecules would seprte the two ions. For concentrted solutions the ctivity would be much less thn the concentrtion itself. Therefore, the erlier given expression of ph cn be modified for dilute solutions s, ph = log [H + ]. This ssumption cn only be mde when the solution is very much dilute, i.e, [H + ] M. For higher concentrtion of H + ions, one needs to clculte the ctivity experimentlly nd then clculte the ph. Exercise :.Clculte the ph of 0 M HCl solution Strong Acids Let us now see how to clculte the ph of solution of strong cid in wter (it should be noted tht ph clcultions re only mde for queous solutions). Let the strong cid
be HCl. If we tke 0 M HCl, the [H + ] would be 0 M, s HCl is strong cid nd would dissocite completely. Therefore the ph would be, ph = log 0 = Concentrtion of HCl 0 M 0 M 0 M 0 4 M 4 0 5 M 5 0 6 M 6 ph 0 7 M 7 (?) We cn see tht for 0 7 M of HCl we hve some hesittion in tlking bout the ph. This is becuse if we use our expression of ph, it works out to be 7 which is somehow ssocited with neutrlity. We shll now explin how to clculte the ph of 0 7 M HCl. Before we do this we shll discuss the dissocition of wter. Dissocition of Wter nd Ionic Product of Wter Wter behves s wek cid (or wek bse) s it wekly dissocites to give H + ions (or OH ). H O H + + OH This wek dissocition of wter mkes it rech equilibrium with its ions. H OH = The concentrtion of wter before dissocition is 55.56 moles per liter. Since the dissocition of wter is so feeble tht it cn be ssumed tht the concentrtion of wter t equilibrium is lmost the sme. Therefore we cn ssume tht [H O] is constnt. H O [H O] = [H + ] [OH ] = C We cn see tht the bove expression is C. In this topic we shll give specil sub scripts for different types of C s. Hence the bove expression's which is the product of the concentrtions of H + nd OH is clled the ionic product of wter, represented s w. At 5 C the vlue of w is pproximtely equl to 0 4. [H + ] [OH ] = 0 4 () As wter gives the sme mount of H + nd OH nd if we represent the concentrtion of H + s xm then the concentrtion of OH will lso be xm. x x = 0 4
x = 0 4 () x = 0 7 M () ph = log 0 7 = 7 Now we revert bck to our originl problem which is the clcultion of ph of solution, 0 7 M HCl. When we dd 0 7 M HCl in wter, the [H + ] from HCl would be0 7 M. But ph is the negtive logrithm of the totl H + ion concentrtion of solution nd not tht prt which comes from only HCl ( you might wonder why we hve chnged our stnd. This will become cler in short while). Therefore, we need to dd the concentrtion of H + ions coming from wter lso. This cn be done in two wys. Method I We lredy know the concentrtion of H + ions from pure wter. It is 0 7 M. So ll we need to do is to dd the H + ion concentrtion from wter nd tht from HCl to get the totl H + ion concentrtion. [H + ] T = [H + ] HCl + [H + ] wter = 0 7 M + 0 7 M = 0 7 M ph = log [H + ] T = log 0 7 = 6.6989 Though everything looks O, there is big mistke we hve mde. In Lesson 4 (Chemicl Equilibrium) we hve lernt in Le Chtlier s Principle tht when concentrtion of rectnt or product is chnged the rection tends to go forwrd or reverse to finlly rech equilibrium. We cn see tht in this cse by dding 0 7 M HCl to wter, we re incresing the concentrtion of H + ions. This should mke the rection go in the reverse direction. Therefore the bove given method is incorrect. Method II The sttement given bove tht the ddition of n ion to n equilibrium, hving the sme ion mkes the equilibrium rection move in direction to consume tht ion is clled the common ion effect. This implies tht wter would dissocite less in the presence of HCl. Let the mount to which wter disssocites be x in the presence of 0 7 M HCl. H O H + + OH At equilibrium: x + 0 7 x
[H + ] [OH ] = 0 4 (x + 0 7 ) (x) = 0 4 Clculting for x, we get x = 0.68 0 7 M. One cn clerly see the common ion effect in ction. Wter which ws dissociting to give 0 7 M H + ions, hs now experienced the common ion effect nd hs finlly yielded 0.68 0 7 M H + ions. Therefore, [H + ] T = 0 7 + 0.68 0 7 =.68 0 7 M ph = log (.68 0 7 ) = 6.790 Now, lets nswer the question s to why we tke the H + ions of wter into ccount for clculting the ph of 0 7 M HCl while we never considered it for clculting the ph of 0 6 M, 0 5 M, 0 4 M, 0 M, 0 M nd 0 M HCl. It cn be seen tht the H + ions from wter hs decresed due to the common ion effect. Greter the concentrtion of the common ion dded, greter will be the effect. Therefore for concentrtions higher thn 0 6 M (nd inclusive of 0 6 M), the H + from wter will be even less thn 0.68 0 7 M nd would be so smll in comprison to the [H + ] from HCl, tht we cn ignore the contribution from it. So, finlly we conclude tht H + ions of wter needs to be considered only if n cid is present such tht [H + ] cid < 0 6 M. Illustrtion : Clculte the ph of 0 5 M HCl solution, if ml of it is diluted to 000 ml Solution: [HCl] = 0 8 M, in diluted smple Since this is very low, we must consider [H + ] from dissocition of H O. H O H + + OH X x w = [H + ] [OH ] = 0 4 [H + ] (finl) = [H + ] cid + [H + ] wter = 0 8 + x w = (x+0 8 ) x = 0 4 x + 0 8 x 0 4 = 0 6 4 8 0 4 0 x = 0 x = 9.5 0 8 [H + ] finl = 0 8 + 9.5 0 8 =.05 0 7 ph = log [.05 0 7 ] = 6.978 The sme logic pplies for bses lso. The only difference is tht for deciding whether we should tke the OH ion from wter or not, we should see whether [OH ] from the bse is less thn 0 6 M or not. Exercise: Clculte the ph of pure wter t 5C. Given enthlpy of dissocition of wter is 50.88 J. Wek Acids
Let us tke wek cid (CH COOH) nd see how to clculte its ph. Let its initil concentrtion be C moles / l. CH COOH CH COO + H + Initil C 0 0 At eq. C( ) C C CHCOO H C = CH COOH This equilibrium constnt, C is given symbol. CHCOO H = CH COOH = C C C = C Generlly for wek cids, (degree of dissocition) is very much less thn. Therefore, = C = C We cn see tht s concentrtion decreses, increses. [H + ] = C = C Illustrtion : Clculte the ph of 0 6 M CHCOOH. (CHCOOH) =.8 0 5. Solution: We solve this problem by two methods. Method : Since [H + ] due to wek cid is given by [H + ] = C = ph = 5.7 5 6.8 0 0 = 4.4 0 6 Though we seemed to hve solved it correctly, there is n error tht we hve mde. This error cn be highlighted by considering the ph of 0 6 M HCl. We cn see tht the ph would be 6. Now, we know tht lesser ph implies higher concentrtion of H + ions. So how cn wek cid hving the sme concentrtion s strong cid dissocite to give the sme concentrtion of H + ions. The mistke hs occurred in the ssumption tht is very much smll compred to. In fct by mking this ssumption the tht we clculte is, =.8 0 = 6 C 0 5 = 4.4 This vlue is not possible s cnnot be more thn, ever. Therefore, we clculte the ph without this ssumption.
Method : = C,.8 0 5 = 6 0 Solving for, = 0.95 [H + ] = C = 0.95 0 6 = 9.5 0 7 ph = 6.0 Of course we need to tke the [H + ] from wter lso s [H + ] cid < 0 6 M. (x + 9.5 0 7 ) x = 0 4 (Where x is the mount to which wter dissocites in the presence of 0 6 M CH COOH). x =.04 0 8 [H + ] T = 9.5 0 7 +.04 0 8 ph = 6.0 This illustrtion shows need to know s to when is the ssumption tht is very smll compred to, vlid. This is done in the following mnner. First of ll clculte the vlue of from the expression = C If this vlue of comes out to be less thn or equl to 0., then the ssumption is vlid. If the vlue of is such tht, >0., then the ssumption is not vlid nd one hs to clculte using the qudrtic expression. Exercise: ) Wht is the ph of 7.00-8 M cetic cid? Wht is the concentrtion of unionized cetic cid? b) Clculte [H + ] in solution tht is 0. M HCOOH nd 0. M HOCN. (HCOOH)=.80-4, (HOCN)=.0-4 Hydrolysis of Slts Slts re the product of n cid nd bse, other thn wter. Depending on the nture of n cid or bse there cn be four types of slts: i) Slt of wek cid nd strong bse, ii) Slt of strong cid nd wek bse, iii) Slt of wek cid nd wek bse nd iv) Slt of strong cid nd strong bse. We shll first look t wht is hydrolysis nd then find out how to clculte the ph due to it.
Slt of Wek Acid nd Strong Bse Let us tke certin mount of wek cid (CH COOH) nd dd to it the sme mount (equivlents) of strong bse (NOH). They will rect to produce CH COON. CH COOH + NOH CH COON + H O CH COON being strong electrolyte, completely dissocites into its constituent ions. CH COON CH COO + N + Now, the ions produced would rect with H O. This process is clled hydrolysis. N + + CH COO + H O CH COOH + NOH We know tht NOH is strong bse nd therefore it would be completely dissocited to give N + nd OH ions. N + + CH COO + H O CH COOH + OH + N + Cnceling N + on both the sides, CH COO + H O CH COOH + OH We cn note here tht ions coming from strong bses do not get hydrolysed. We should note here tht the solution will be bsic. This is becuse the mount of CH COOH produced nd OH - produced re equl. But CH COOH will not completely dissocite to give H + ions. Therefore [OH ] ions will be greter thn [H + ] ions. Since the rection is t equilibrium, C = CH COOH OH CH COO This equilibrium constnt c is given new symbol, h. h CH COOH OH [CH COO ] If we multiply nd divide the bove eqution by [H + ] of the solution, then CHCOOH OH H h H OH = = [CH COO ] [H ] [CHCOO ] [H ] [CH COOH]
h = w CHCOOH OH h = [CH COO ] = w CH COO +H O CH COOH + OH Initil: C 0 0 At eq: C( ) C C Where is the degree of hydrolysis of CH COO ion. C C C = w If is very much less thn, C = w, = w C As [OH ] = C, [OH ] = C w C w C w [H + ] = w C OH or ph = log [H + ] = log W C = p W p logc Now, we will nlyse the hydrolysis of this slt in detil. The rections occurring were, CH COOH + NOH CH COON + H O CH COON CH COO + N + () () CH COO + H O CH COOH + OH () If we look t the first rection, we cn see tht NOH will be fully ionized s N + & OH ions nd CH COON s CH COO & N + ions. Cnceling N + from both the sides, the first rection looks like, CH COOH + OH CH COO + H O (4) You cn see clerly tht rection (4) nd () re the reverse of ech other. Therefore, rection (4) is lso in equilibrium. CH COOH + OH CH COO + H O (5)
This ctully puts us in dilemm. If we look t rection (5) [which is bsiclly the reverse of rection ()], we see tht when CH COOH rects with OH the rection reches equilibrium. Now, there re two possibilities: (i) either cetic cid rects with OH nd the rection reches equilibrium or (ii) cetic cid rects with OH nd the rection completely goes to produced CH COO which then gets hydrolyses (by going in the reverse direction) nd finlly reches equilibrium. On close scrutiny we would relize tht the first possibility is more relistic. It is not logicl for rection to completely go to the right nd then go reverse to rech equilibrium. Now, if the first explntion is correct then we find tht ctully no hydrolysis hs tken plce. Wht hs hppened is tht the first rection hs not gone to completion nd hs reched equilibrium. Aprt from this difference, both the explntion would give us the sme result in terms of ph clcultion. For exmple, we cn see tht if certin mount of CH COOH rects with the sme mount (equivlents) of OH ions nd the rection reches equilibrium, then equl mounts (equivlents) of OH ions nd CH COOH would be left over. This is similr sitution which we encountered erlier in hydrolysis. We cn see tht this solution would be bsic. To conclude we cn sy tht hydrolysis of slt occurs only when the slt is dissolved in wter. Illustrtion : Clculte the ph fter the ddition of 90 ml nd 00ml respectively of 0.N NOH to 00ml 0.N CHCOOH (Given p for CHCOOH = 4.74) Solution: If 90 ml. Of.N NOH is dded to 00 ml. Of.N CH COOH, cidic buffer will form s H CCOOH + NOH H CCOON + H t = 0.0 eq..009 eq. 0 0 ph = p + log = 4.74 + log.00 eq. 0.009 eq. 9 [CHCOO ] [CH COOH] = 4.74 +.954 = 5.694 If 00 ml of 0.N NOH is dded to 00 ml of 0.N CH COOH, complete neutrlistion tkes plce nd the concentrtion of H CCOON 0. = M =.05M Now, ph = 7 p log C ph = 5.8 Slt of Wek Bse nd Strong Acid Let the cid be HCl nd the bse be NH 4OH. Therefore the slt would be NH 4Cl.
NH 4Cl completely dissocites into NH 4Cl Cl + NH 4 NH 4 +H O + Cl NH 4OH + HCl NH 4 nd Cl ions. HCl being strong cid dissocites completely to give H + ions nd Cl ions. Cl + NH 4 NH 4 + H O NH 4OH + H + + Cl + H O NH 4OH + H + We cn see tht the ion coming from the strong cid does not get hydrolysed. We hd previously seen tht the ion coming from the strong bse lso does not get hydrolysed. Hence, one cn conclude tht the slt of strong cid nd wek bse does not get hydrolysed. In this hydrolysis, NH 4OH nd H + re being produced. This implies tht the solution is cidic. To clculte ph, NH 4 + H O NH 4OH + H + Initil: C 0 0 At eqb: C( ) C C Where is the degree of hydrolysis of NH4OH H h = [NH 4 ] NH 4 Multiplying nd dividing by OH nd rerrnging, NH 4OH H OH h = 4 [NH ] [OH ] H OH = h = [NH 4 [NH w b ] [OH 4 OH] ] = Now, substituting the concentrtions, NH OH 4 w h = C [NH ] H If 0., then, C = = w b C 4 b w b w b C C C.
Since [H + ] = C, [H + ] = C b w C = wc b or ph = p W p b logc Slt of Wek Acid nd Wek Bse Let the wek cid be CH COOH nd the wek bse be NH 4OH. Therefore, the slt is CH COONH 4. The slt completely dissocites. CH COONH 4 CH COO + NH 4 The ions get hydrolysed ccording to the rection. CH COO + NH 4 + H O NH 4OH + CH COOH Initil : C C 0 0 At equilibrium: C( ) C( ) C C NH 4OH CHCOOH h = 4 [NH ] [CH COO ] Multiplying nd dividing by H + & OH nd rerrnging, h = NH OHCH COOH H OH 4 4 [NH ] [CH COO ] [H ] [OH ] H OH [CHCOO ] [H ] [NH4 ] [OH ] [CH COOH] [NH OH] 4 w b h = w b Substituting the concentrtion terms, h = w b C C C C C C w b There is n importnt issue tht needs clrifiction before we move on further. In this cse, we cn see tht both the ions (i.e, ction nd nion) get hydrolysed to produce wek cid nd wek bse (hence, we cn t predict whether the solution is cidic, bsic or neutrl). We hve considered the degree of hydrolysis of both the ions to be the
sme. Now we present n explntion s to why this is incorrect nd then stte resons for the vlidity of this ssumption. Actully the hydrolysis rection given erlier, CH COO + NH 4 + H O CH COOH + NH 4OH is mde up of the following three rections, CH COO + H O CH COOH + OH NH 4 + H O NH 4OH + H + H + + OH H O If we dd these three rections, the net rection is the one stted previously. This suggests tht both CH COO nd get hydrolysed independently nd their NH 4 hydrolysis depends on: (i) Their initil concentrtion nd (ii) the vlue of h which is for CH COO nd w b for NH 4. Since both of the ions were produced from the sme slt, their initil concentrtions re sme. Therefore unless nd until the vlue of nd w b or nd b is sme, the degree of hydrolysis of the ions cnnot be sme. To explin why this ssumption is vlid, we need to now look t the third rection, i.e., combintion of H + nd OH ions. It is obvious tht this rection hppens only becuse one rection produced H + ion nd the other produced OH ions. We cn lso note tht this rection cuses both the hydrolysis rection to occur more since their product ions re being consumed. If you look t the solution of Exercise, you will notice tht the equilibrium which hs smller vlue of equilibrium constnt is ffected more by the common ion effect. For the sme reson if for ny reson rection is mde to occur to greter extent by the consumption of ny one product ion, the rection with the smller vlue of equilibrium constnt tends to get ffected more. Therefore, we conclude tht firstly the hydrolysis of both the ions occurs more in the presence of ech other (due to the consumption of the product ions) thn in ech other s bsence. Secondly the hydrolysis of the ion which occurs to lesser extent (due to smller vlue of h) is ffected more thn the one whose h is greter. Hence we cn see tht the degree of hydrolysis of both the ions would be close to ech other when they re getting hydrolysed in the presence of ech other. Now, to clculte the ph of this solution, we need to understnd one more principle. Let us ssume tht we hve n queous solution of CH COOH. Let us lso ssume tht we do not know whether CH COO nd H + re in the solution or not (even though H + would be there since the solution is queous). We cn see tht the Q for the rection w w
CH COOH CH COO + H + is zero (if we ssume no CH COO or H + ). Q <. The rection would therefore move forwrd to rech equilibrium. The sme is true when CH COO nd H + re present without CH COOH. Hence, we conclude tht Any equilibrium tht cn exist (i.e., when ll rectnts nd / or ll products re present) will exist (except when pure solids or liquids re not present, in which cse this sttement will not be true). As cn be seen in the hydrolysis rection, CH COO + NH 4 + H O NH 4OH + CH COOH, CH COOH is present in the solution. This implies tht the equilibrium between CH COOH, CH COO nd H + cn exist nd therefore would exist. CH COOH CH COO + H + In fct the equilibrium between NH 4OH, NH 4 Now, we clculte the ph of the solution s, CH COOH CH COO + H + C CHCOO H = CH COOH C( ) = H C C nd OH lso exists. [H + ] = Substituting s W b [H + ] = = w b = w b or ph = p W p p Buffer Solutions b A buffer solution is solution which resists chnge in its ph when such chnge is cused by the ddition of smll mount of cid or bse. This does not men tht the ph of the buffer solution does not chnge (we mke this ssumption while doing
numericl problems). It only mens tht the chnge in ph would be less thn the ph tht would hve chnged for solution tht is not buffer. There re three types of buffer solutions: i) wek cid slt buffer ii) wek bse slt buffer nd iii) slt buffer Buffer of Wek Acid nd its Slt with Strong Bse It is possible to prepre buffer solution by the ddition of wek cid nd slt of the cid with strong bse. We shll explin the buffer ction by the following exmple. Let us consider buffer solution mde up of CH COOH nd CH COON. The wek cid dissocites to very smll extent more so due to the common ion effect of its slt. CH COOH CH COO + H + CH COON CH COO + N + Now let us ssume tht buffer solution contins 0 moles of CH COOH nd 0 moles of CH COON. The slt being strong electrolyte would completely dissocite while the cid would be hrdly dissocited. We ssume tht the mount of CH COO is 0 moles s the contribution from CH COOH would negligible. We lso ssume tht the mount of CH COOH to be 0 moles s it would be very wekly dissocited. Therefore the solution contins 0 moles of CH COOH nd 0 moles of CH COO ions. Let us now dd 0 moles of H + ions to this solution. These 0 moles of H + ions would rect with the 0 moles of CH COO ions to produce 0 moles of CH COOH. (This is becuse the rection of CH COOH to give CH COO ions nd H + ions hs n equilibrium constnt vlue of pproximtely 0 5. Therefore, the reverse rection, tht is the rection of CH COO ions to combine with H + ions to give CH COOH would hve n equilibrium constnt of pproximtely 0 5. So this rection cn be ssumed to be complete). These 0 moles of CH COOH formed would ionize wekly becuse it is wek cid. Moreover due to the common ion effect of CH COO ion (0 moles, left over), it would ionize even less. So the mount of H + ion produced bck is much less thn the 0 moles tht were dded. So effectively, 0 moles of H + ions were consumed nd n mount much less thn tht is produced bck which cuses n insignificnt chnge in ph of the buffer solution. Let us dd 0 moles of OH ions to the sme buffer solution contining 0 moles of CH COOH nd 0 moles of CH COO ions. These 0 moles of OH ions dded would be consumed by the 0 moles of CH COOH to produce 0 moles of CH COO ions (CH COOH + OH CH COO + H O. This is becuse the reverse rection is the
hydrolysis of CH COO ions which hs n equilibrium constnt of w 0 9. So the forwrd rection s equilibrium constnt would be 0 9, which implies tht the rection is prcticlly complete). The 0 moles of CH COO ions produced would be hydrolysed wekly (due to very low vlue of 0 9 ). On top of tht, the presence of 0 moles of CH COOH would further hmper the hydrolysis process nd thereby the OH ion produced is much less thn the mount tht ws dded. This cuses minor chnge in ph. For buffer to ct s good buffer the mount of CH COOH nd CH COO ion should be high. The ph of buffer cn be clculted s follows. Since cetic cid is in equilibrium, [H ] [CHCOO ] = [CH COOH] [H ][Conjugte bse] = [H + ] = [Acid] log [H + ] = log + log log [H + ] = log + log ph = p + log [Acid] [Conjugte bse] [Conjugte bse] [Acid] [Conjugte bse] [Acid] [Acid] [Conjugte bse] This eqution is clled Henderson Hsselblch eqution. When wek dibsic cid is tken then the ph t hlf neutrlistion is given by specil formul which is derived by blncing chrge nd mss of rectnts nd products. For exmple, t hlf neutrlistion of H CO by NOH, the ph is given by p p ph = Where nd re st nd nd dissocition constnt of H CO. Buffer of Wek Bse nd its Slt with Strong Acid This combintion works on the principle s stted in the previous section. If we tke the exmple of NH 4OH nd NH 4Cl, then the OH ion dded would rect with NH (from the slt) to produce NH 4OH. The H + ion dded would rect with NH 4OH to produce The Henderson Hsselblch eqution ppers s 4 NH 4.
Conjugte cid poh = p b + log Bse Illustrtion 4: CHCOOH (50 ml, 0.M) is titrted ginst 0.M NOH solution. Clculte the ph t the ddition of 0 ml, 0 ml, 0 ml, 5 ml, 40 ml, 50 ml nd 60 ml of NOH. of CHCOOH is 0 5. Solution: i) When 0 ml of NOH is dded, the ph is due to cetic cid, [H + ] = ph = log C 0 = 0 5 0. = 6 = 0 6 [log 6] = 0.5 =.85 ii) When 0 ml of NOH is dded, it rects with CH COOH to produce slt nd wter. The solution is then buffer. Conjugte bse ph = p + log Acid = 4.699 + log 0 0. = 4.699 + log 50 0. 0 0. 4 = 4.0969 iii) When 0 ml of NOH is dded. 0 0. ph = p + log = 4.699 + log 50 0. 0 0. = 4.59 iv) When 5 ml of NOH is dded, 5 0. ph = 4.699 + log 50 0. 5 0. = 4.699 v) When 40 ml of NOH is dded, 40 0. ph = 4.699 + log = 4.699 + log 4 = 5.0 50 0. 40 0. vi) When 50 ml of NOH is dded, Here, if we use the buffer eqution, ph would be = But we cn t use the buffer eqution s there is no cid. Therefore we used the hydrolysis eqution (refer to illustrtion ) [H + W ] = C 0. C = [ Totl Volume is 00 ml nd millimoles of slt is 50 0.] [H + ] = 0 4 0 0. 5 = ph = 8.699 vii) When 60 ml of NOH is dded, the excess OH ion from NOH would suppress the hydrolysis of CH COO ion. So we cn ignore the contribution of OH ion from the hydrolysis of CH COO ion.
[OH ] = poh =.044 0.0 0 ph = 4.044 =.9586 [ 0 ml of OH ion is in excess] = 0 Indictors In the illustrtion, we hve clculted the ph during the titrtion of wek cid with strong bse t vrious stges. If we clculte the ph t severl instnces during the titrtion nd then plot grph between the ph nd the volume of strong bse, we get curve s shown in figure. 8.69 6 ph 49.9ml 50 ml strong bse The curve hs been drwn by considering the titrtion of 50 ml of 0.M CH COOH with 0.M NOH. Let us first of ll understnd the nture of the curve. Initilly the ph chnges rpidly with the volume of the bse nd then its chnge becomes very slow nd gin it chnges rpidly. This phenomenon cn be explined like this. Initilly, strong bse is dded to wek cid. Therefore the ph chnges ccordingly. But fter the ddition of certin mount of the bse, the solution turns into buffer s contins CH COOH nd CH COON. So ny further ddition of NOH cuses very little chnge in ph. Therefore the slow chnge s more nd more of strong bse is dded, the slt concentrtion increses nd the cid concentrtion decreses nd especilly ner the equivlence point, the cid concentrtion is very smll. When there is such lrge difference in their concentrtion, the solution stops cting s buffer nd there is sudden jump in the ph. Now, let us ssume tht when 49.9 ml of NOH hs been dded the ph of the solution is 6 ( this is just n ssumption to illustrte the role of indictors nd not the ctul vlue) nd when 50 ml of NOH is dded (equivlence point) the ph is 8.69. If we ssume tht one drop is equl to 0. ml then wht we hve is sitution in which on dding the lst drop of NOH the ph chnges from 6 to 8.69. An idel indictor would be the one which chnges its colour on the ddition of the lst drop of NOH. Let the indictor be wek cid HA (ll cid bse indictors re either wek cids or wek bses). It would dissocite wekly s, HA H + + A. Let the wek cid hve colour red in cid solution nd hve colour blue in bsic solution. This implies tht HA is red (s HA would be lmost undissocited in cidic solution) nd A is blue (s HA would be lmost completely dissocited in bsic solution). Now,
ind = [H ] [A ] [HA] [H + [HA] ] = ind [A ] where ind is the of the indictor. () To see ny colour, we need to hve certin minimum concentrtion of the coloured species. Let the minimum mount of HA required to be present to mke the colour red visible be 90% of the totl indictor. This mens tht if the indictor hs HA, 90% or more in mount, then we would see the colour red nd if it is less thn 90% then coloured red would not be visible. Similrly we cn ssume tht to see the colour blue, the minimum mount of A required to be present is lso 90%. This mens tht if 90% or more of the totl indictor is A, then colour blue would be visible, otherwise not. We cn see tht the minimum rtio of [HA] [A ] to see the red colour would be 9. Let the [H + ] t this instnt (from eqution ) be 0 6.5. This is therefore, the minimum mount of H + required to see the red colour. Converting this to ph ( log [H + ]), we get the mximum [HA] ph to see the red colour (i.e., 6.5). Similrly, the mximum rtio of to see the blue [A ] colour would be 9. Let the [H + ] t this instnt (from eqution ) be 0 8. This is therefore the mximum mount of H + required to see the blue colour. Converting this to ph ( log [H + ]), we get the minimum ph to see the blue colour (i.e., 8). Hence the indictor would chnge colour between ph 6.5 8. It would be red t ph 6.5 or below. It would be blue t ph 8 or bove. Between the two ph vlues, it would be colourless. An indictor would be effective for given titrtion if its rnge of ph for colour chnge (which in this cse is between 6.5 to 8) is within the rnge of ph before nd fter the lst drop of n cid or bse is dded (which is in this cse is between 6 to 8.69). We cn esily see tht if this prticulr indictors is dded in the solution, it would hve red colour before the ddition of the lst drop nd blue colour fter the ddition of the lst drop. Exercise 4: Distinguish between the end point nd the equivlence point of titrtion, explin using these two terms the importnce of choosing the proper indictor for given titrtion. Slt Buffer The fundmentl principle behind buffer ction is the fct tht on dding n cid the system consumes the H + ion dded to produce wek cid nd on dding bse, it consume the OH ion dded to produce wek bse. This ensures tht H + or OH ion
dded is consumed nd the wek cid or the wek bse produced gives less H + or OH ion s they re wek. Bsed on this principle, solution of slt of wek cid nd wek bse is lso clled s buffer. Let us tke the exmple of CH COONH 4. It dissocites s, CH COONH 4 CH COO +. When H + ion is dded, CH COO ion consumes it NH 4 to give CH COOH. When OH ion is dded, it cts s buffer. [H + ] = w b (derived erlier). NH 4 Solubility nd Solubility Product ion consumes it to give NH 4OH. Hence Here we discuss the solubilities of ionic solutes. When slt is dissolved in wter, it dissolves s the slt, AgCl (s) AgCl(q) nd then the dissolved slt dissocites to give the ions AgCl (q) Ag + (q) + Cl (q). After certin mount of the slt hs been dissolved, the solution would become sturted with the slt. Now, if some more of the slt is dissolved, the slt would dissolve to give ions while t the sme time the ions would precipitte to give bck the slt. This mount (moles) of the slt tht hs mde the solution sturted per liter of solution is clled the solubility of the slt. AgCl (s) Ag + (q) + Cl (q) = [Ag ] [Cl ] [AgC] Since the concentrtion of AgCl, which is solid, is constnt, [AgCl] = [Ag + ] [Cl ] = C This C is given the symbol, SP. This mens tht solution cnnot hve the product of the concentrtion of Ag + nd Cl to be more thn SP (clled solubility product). Clcultion of Solubilities of Slts We shll now discuss the solubilities of different types of slts under vrious conditions. i) Solubilities of AgCl (slt of strong cid nd strong bse) in wter: AgCl would dissolve in wter s, AgCl (s) Ag + (q) + Cl (q) At sturtion point, AgCl (s) Ag + (q) + Cl (q) If the solubility of the slt is x moles / l [Ag + ] = xm, [Cl ] = xm
x = sp x = sp ii) Solubility of AgCl in solution tht is hving 0.M in AgNO : AgCl would dissolve nd finlly rech sturtion. AgCl(s) Ag + (q) + Cl (q) The sp of AgCl is pproximtely 0 0. If AgCl were to be dissolved in wter (pure), its solubility would hve been 0 5 M (previous section). In the presence of 0.M AgNO its solubility will decreses due to common ion effect. This mens tht [Ag + ] from AgCl would be less thn 0 5 M. Hence, we cn ignore the contribution of Ag + from AgCl. If the solubility of AgCl is x moles / l in the presence of 0.M AgNO, then [Ag + ] = 0.M, [Cl ] = x M x = sp 0. = 0 9 moles / l iii) Solubility of CH COOAg (slt of wek cid nd strong bse) in wter: CH COOAg dissolves nd reches sturtion. Since it is slt of wek cid nd strong bse, it would hydrolyse. If the solubility of the slt is x moles/ l then CH COOAg (s) CH COO (q) + Ag + (q) At eqb: x y x CH COO (q) + H O CH COOH (q) + H + (q) At eqb: x y y y Where y is the mount of CH COO ion tht is hydrolysed. (x y) x = sp y x y = w nowing the vlues of sp nd, solubility of the slt cn be clculted. iv) Solubility of CH COOAg (slt of wek cid nd strong bse) in n cid buffer of ph = 4 (ssuming tht the buffer does not hve ny common ion by CH COOAg): CH COOAg would dissolve nd rech equilibrium. It would then be hydrolysed. If the solubility of the slt is x M in this solution, then
At eq.; CH COOAg (s) CH COO (q) + Ag + (q) x y x CH COO (q) + H + CH COOH (q) x y 0 4 y Since the solution is buffer, the ph will be mintined. ( x y) x = sp 0 y x y 4 = Since in presence of bsic buffer, the degree of hydrolysis will be suppressed by lredy existing OH ions, therefore the pproximted formul which cn be used is X = sp (CH COOAg) (neglecting y) nowing sp nd, the solubility cn be clculted. v) Solubility of CH COOAg in n buffer solution of ph = 9: Following the sme logic s give in the erlier section, CH COOAg (s) CH COO (q) + Ag + (q) At eqb: x y x CH COO (q) + H O CH COOH + OH At eqb: x y y 0 5 Where x M is the solubility of the slt nd y the extent to which it is hydrolysed. ( x y) x = sp 5 y 0 (x y ) = w nowing, sp nd, the solubility cn be clculted. vi) Solubility of AgCl in n queous solution contining NH. Let the mount of NH initilly be M. If the solubility of the slt is x moles/ l, then AgCl (s) Ag + (q) + Cl (q) At eq: x y x Ag + (q) + NH (q) Ag(NH ) + (q) x y y y Where y is the mount of Ag + which hs rected with NH. ( x y) x = sp
y x y y = f ( formtion constnt of Ag(NH ) + ) nowing sp nd f, the solubility cn be clculted. Precipittion of slts For slt (springly soluble) when dissolved in wter : AB (sturtion) sp = [A + ] [B ] A + + B t equilibrium When, we mix ions or if there be two or more ions in wter, we define rection coefficient (Q), clled s ionic product (IP) in this cse, giving the products of ions in wter (ions of soluble slts nd other common ions). I.P. is product of ionic concentrtion due to ions lredy present in wter or from slt. I.P. my be nd my not be equl to sp. To illustrte it more clerly, consider cse when 500 ml of n 0.005 M solution of AgNO is dded to 500 ml of 0.00 M solution Cl. Now in solution (mixture), there re Ag +,, + nd Cl ions. The concentrtion NO of [Ag + ] = NO = 0.005 = 0.005 M (equl volumes re mixed) nd [ + ] = [Cl ] = = 0.0005 M s equl volumes of two solutions re mixed. Now, we know tht Ag + will rect with Cl ions to form AgCl since: 0.00 AgCl Ag + + Cl is reversible rection with higher tendency towrds left (solidifying or precipitting). Now, question is, whether AgCl will be formed or not (precipittion of Ag + nd Cl s AgCl) nd if it formed, how much of it will be formed? For this we define some rules.. If ionic Product (IP) sp ; precipittion tkes plce till I.P. equls sp. If Ionic Product sp ; precipitte will not be formed nd the solution will be unsturted. If Ionic Product = sp ; precipitte will not form n the solution is stuted in tht slt. (or we cn sy tht solution is t criticl stge, when precipittion just begins, but ctully hs not occurred yet in rel sense). In present cse, I.P. = [Ag + ] [Cl ] = (0.005) (0.0005) =.5 0 6 (Only for the slt which is spring soluble not for NO ) Ionic Product sp in this cse ( sp AgCl =.56 0 0 ) Which men precipittion tkes plce.
Illustrtion 5: The solubility of CF in wter t 0C is 5.6 mg per dm of solution. Clculte the solubility product of CF. Solution: Solubility in moles per dm 5.6 0 = (78 g / mol) =.0 0 4 Since CF C + + F Therefore, [C + ] =.0 0 4 nd [F ] =.0 0 4 Hence, solubility product sp = [C + ] [F ] = (.0 0 4 ) (4.0 0 4 ) = 0 Illustrtion -6: The solubility of BSO4 in wter is. 0 4 gm/00 ml. Clculte the percentge loss in weight when 0. gm of BSO4 is wshed with () lt of wter (b) lt of 0.0 NHSO4. Solution: ) Solubility is in generl expressed in gm/lt, so solubility of BSO 4 =. 0 g /lt Loss in weight of BSO 4 = mount of BSO 4 soluble %loss =. 0. b) Now 0.0 NH SO 4 0.0 0.005 M SO 4 ions 00 =.5% NSO 4 ions Now presence of SO 4 prior to wshing BSO 4 will suppress the solubility of BSO 4 (due to common ion effect). The suppression will be governed by sp vlue of BSO 4. So first clculte sp of BSO 4. Solubility of BSO 4 in fresh wter = 0 g/lt. 0 = Mol/lt = 9.85 0 6 M.4 sp = [B + ] SO = (9.85 0 6 ) = 9.7 0 4 Now let x be solubility in mol/lt [B + ] in solution = x mol/lt nd SO in solution = (x + 0.005) mol/lt Ionic product = [B + ] SO = (x) (x + 0.005) sp = Ionic Product t equilibrium (sturtion) 4 9.7 0 = (x) (x + 0.005) x + 0.005 x - 9.7 0 = 0 4
x = 9.70 0.005 =.94 0 8 mol/lt =.94 0 8.4 g/lt 4.5 0 6 gm of BSO 4 re wshed wy percentge loss = 4.5 0 0. 6 00 =.6 0 % ph of Mixtures of Acids & Bses Let us tke x millimoles of cid (HA) nd y millimoles of bse (BOH). Note tht cid is monobsic nd bse is monocidic. ) Strong cid & Strong bse: If x = y; then complete neutrliztion tkes plce nd we get x(=y) millimoles of slt (BA) of strong cid nd strong bse which mens no hydrolysis tkes plce nd ph of solution = 7. If x y; then there is n excess of strong cid nd resulting solution is cidic with millimoles of cid left in excess = x y. Now if V c.c. be the volume of mixture, then: M = x y. Now clculte ph using the eqution ph = log [H+ ] V If x y; then there excess of strong bse nd resulting solution is bsic with millimoles of bse = y x. b) Strong bse & wek cid If x = y; first of ll neutrliztion tkes plce to give x(=y) millimoles of slt (BA). The slt will now undergo hydrolysis to give n lkline solution. Clculte ph by using stndrd result: ph = 7 + (p + log C); C is concentrtion expressed in M (mol/lt) If x y; there is excess of wek cid whose millimoles = x y nd y millimoles of slt is formed. This will give n cidic buffer solution. Clculte ph of buffer solution using Henderson s Eqution. PH = p + log [slt] [cid] p + log x y y If x y; the solution in this cse contins excess of strong bse whose millimoles re y x.
M = y x. Clculte ph. V c) Strong cid & Wek bse If x = y; first of ll complete neutrlistion tkes plce to produce x(=y) millimoles of slt (BA). The slt (BA) is of strong cid nd wek bse, hence hydrolysis tkes plce to give n cidic solution. Clculte its ph by using stndrd result. ph = 7 (pb + log C) ; C: is concentrtion of slt. If x y; then solution contin excess of strong cid whose millimoles = x y. M = x y V. Clculte ph. If x y; then there is n excess of wek bse whose millimoles re y x nd millimoles of slt (BA) re x. This will give bsic buffer solution. Clculte the ph by using Henderson s Eqution. ph = 4 pb log [slt] [bse] d) Wek cid & wek bse = 4 pb log If x = y; neutlistion tkes plce completely with the formtion of x (=y) millimloles of slt (BA) of wek cid nd wek bse. So hydrolysis tkes plce. Clculte the ph by using: ph = 7 + (p p b) If x y; then excess of wek cid (x y) will remin with y millimoles of slt. This will give n slt buffer with cidic ph. If x y; then excess of wek cid (y-x) nd x millimoles of slt. This will give slt buffer with bsic ph. Solutions to Exercise y x x Exercise : solution Becuse ctivity is not known so we cn not clculte the ph of w5 50.88 0 Exercise : log.00 R 98 08 w5 = w(5c) =0 -
[H + ] = 0-6 PH=6 Exercise : ) (In cse of dilute cid soln, we hve to ssume complete ioniztion of the wek cid but we must tke into ccount the ioniztion of wter) H O H O + +OH _ w= [H O + ][OH - ]=.0 0-4 Let, [OH - ]=x [H O + ]=(7.00-8 )+X { from wter nd cid } w=x(x+7.00-8 ) =.0 0-4 or, x +7.00-8 x.0 0-4 =0 x=7.0-8 M = [OH - ] thus, [H O + ]=7.0-8 + 7.00-8 =.40-7 M thus, ph=-log [H O + ]=6.85 for chnge blnce, [CH COO - ]+[OH - ]=[H O + ] or, [CH COO - ]=[H O + ]-[OH - ] =.40-7 -7.0-8 = 70-8 M CH COOH+H O = CH COO - +H O + [ CH COO ][HO ] 5.8 0 [CH COOH] [CH COOH]= (.4 0 7.8 0 )(7 0 5 8 ) = 50-0 M b) This is cse in which two wek cids both contributes to [H + ], neither contributing such big mount tht the others shre cn be neglected. (i) HCOOH HCOO - +H + (x y)(x) initil 0. o o.8 0 (0. x) t eq 0.-x x x-y - (i) 4
(ii) HCON H + +OCN - + (x y)(y) initil 0. 0 0. 0 (0. y) t eq 0.-y (x+y) y - (ii) 4 In eq (i) & (ii), 0.-x 0. 0.-y 0. (Since both re wek cids) From eq (i) & (ii), y x =.8 or y=.8 x..(iii) on subtrction (i) from (ii) y(x y) x(x y) 4.5 0 0. or, y -x =.50-5...(iv) From eq (iii) & (iv), y=4.60 -, x=.50 - & [H + ]=x+y=.50 - + 4.60-7. 0 - M (Ans) Exercise 4: The end point is the point t which the titrtion is stopped. The equivlence point is the point t which the cid nd the bse (or oxidizing gent nd reducing gent) hve been dded in equivlent quntities, since the purpose of the point is to stop the titrtion close to point t which the cid nd bse were dded in equivlent quntities, it is importnt tht the equivlent point nd the end point be s close s possible. The indictor must chnge colour t ph close to tht of solution of the slt of the cid nd bse. Solved Problems Subjective Problem : Clculte the PH of the solution when 0. M CH COOH (50 ml) nd 0. M NOH (50 ml) re mixed, [ (CHCOOH)=0-5 ]
Solution: CH COOH CH COO _ + H + (I) NOH N + + OH _ H + + OH _ H O (II) (I) + (II) CH COOH + OH - CH COO - + H O. (III) 0.05-X 0.05-x x eq of eq. (III) = w w CHCOO = CH COOH OH conc. of H O remin constnt. x 0 9 = 0.05 x becuse vlue of eq. Const.is very high there for x 0.05 let 0.05-x= 0.05 0 9 = = = 5 0 0 9 50 0 6 POH= 6-log 7.07 POH= 6-0.85 PH= 4-6+0.85 =8.85 Problem : Clculte the ph t the equivlence point of the titrtion between 0.M CHCOOH ( 5 ml) with 0.05 M NOH. (CHCOOH) =.8 0 5. Solution: We hve lredy seen tht even though when CH COOH is titrted with NOH the rection does not go to completion but insted reches equilibrium. We cn ssume tht the rection is complete nd then slt gets hydrolysed becuse, this ssumption will help us to do the problem esily nd it does not effect our nswer. [H + ] = w C First of ll we would clculte the concentrtion of the slt, CH COON. For reching equivlence point, N V = N V 0. 5 = 0.05 V V = 50 ml Therefore [CH COON] = 0. 5 75 0.
[H + ] = 0 4 5.8 0 0. ph = log. 0 5 = 8.6 =. 0 5 Problem : Given the solubility product of Pb (PO4) is.5 x 0 -.Determine the solubility in gms/litre Solution: Solubility product of Pb (PO 4) =.5 0 Pb (PO 4) Pb + + PO 4 If x is the solubility of Pb (PO 4) Then sp = (x) (x) = 08 x 5 5 sp 08 5.5 0 x = = 08 x =.69 0 7 moles/lit Moleculr mss of Pb (PO 4) = 8 x =.69 0 7 8 g/lit =.7 0 4 g/lit Solubility product is sp(src O 4) = [Sr + ] [C O 4 ] = (5.4 0 4 ) - =.9 0 7 Problem -4: Wht is ph of M CHCOOH solution? To wht volume must one litre of this solution be diluted so tht the ph of resulting solution will be twice the originl vlue. Given : =.8 0 5 Solution: H CCOOH + H O H CCOO + H O + t = 0 M 0 0 -xm xm xm t = t eq (-x)m x x = x = x x x = 4. 0 = [H O + ] ph = log [H O + ] = log {4. 0 } = log 4. =.7 Now, let L of M ACOH solution be diluted to VL to double the ph nd the conc. of diluted solution be C. H CCOOH + H O H CCOO + H O + t = 0 C 0 0.8 0 5.8 0 5.8 0 5 t=t eq C.8 0 5.8 0 5.8 0 5 New ph = old ph =.7 = 4.74 ph = log [H O + ] = 4.74 [H O + ] =.8 0 5
[CHCOO ] [HO ] = [CH COOH].8 0 5 =.8 0 5.8 0 C.8 0 C =.6 0 5 L on dilution M V = M V M L =.6 0 5 L V V =.78 0 4 L 5 Problem 5: Find the concentrtion of H +, 5 * HCO nd crbonic cid if the ph of this is 4.8 (HCO) = 4.45 0 7 nd = 4.69 0 CO, in 0.0M solution of Solution: ph = log[h + ] 4.8 = log [H + ] [H + ] = 6.6 0 5 H CO H + + [H ][HCO ] = [H CO ] [ HCO 6 5.60 [HCO ] or, 4.45 0 7 =.0 HCO ] = 6.7 0 5 gin, HCO H + + = [ [H ][[CO [HCO ] CO ] = 4.8 0 ] CO = 4.69 0 = 6.60 5 6.7 0 5 [CO ] Problem 6: Clculte the molr solubility of Mg(OH) in MNH4Cl spmg(oh) =.8 0 b(nh) =.8 0 5 Solution: Mg(OH) (s) Mg ++ + OH = sp NH 4 + OH NH 4OH = Add Mg(OH) (s) + NH 4 Mg ++ sp + NH 4OH = b M x x Let molr solubility of Mg(OH) in MNH 4Cl be x.8 0 x (x). = = 4x = = 0.8.8 0.8 8 x =.4 M b
Problem 7: Solution: Problem 8: An queous solution of metl bromide MBr (0.05M) in sturted with HS. Wht is the minimum ph t which MS will ppt.? sp =(MS) = 6 0 Concentrtion of stndrd HS =. (HS) = 0 7 (HS) =. 0 In sturted solution of MS MS(s) M ++ + S [S ] = [M SP 6 0 =. 0 9 ].05 The precipitte of MS will form only if [S ] exceeds the concentrtion of. 0 9 H S H + + HS H S H + + S -- H S H + + S =. 0 0 [H ] [S [ = [H S]. 0 0 = [H + ] =.09 ph =.96 [H ] 9. 0. How much AgBr could dissolve in.0 L of 0.4 M NH? Assume tht [Ag (NH)] + is the only complex formed given, f [Ag(NH) + ]=.00 8, sp (AgBr)= 5.00 - Solution: AgBr Ag + + Br - Ag + + NH Ag (NH ) + Let x= solubility, Then x= [Br - ]=[Ag + ]+[Ag(NH ) + ] [Ag(NH ) ] 8 0 [Ag ][NH ] since, [NH ]>> [Br - ], [Ag(NH ) ].0 0 [Ag ](0.4) 8 (Ag ) 7 [Ag(NH ).6 0 Since, the mjority of silver is in the form of the complex, X=[Br - ] = [Ag(NH ) + ] sp= [Ag + ] [Br - ] [ Ag(NH ) ][Br ] x = 5.0 0 7 7.6 0.6 0 x =8.00-6 x=.80 - M ]
Problem 9: Solution: extent. Problem 0: Equl volumes of 0.0 M Ag NO nd 0.0 M HCN were mixed. Clculte [Ag + ] t equilibrium given, sp (AgCN)=.0-6 (HCN)= 6.0-0 Initilly, ssume complete precipittion Ag + + HCN AgCN + H +, since the solution were diluted to double volume 0.0 concentrtion of [H + ] = 0.0 M Now consider the equilibrium AgCN Ag + + CN - sp =.0-6 = [Ag + ] [CN - ] HCN H + + CN - sp = 6.0-0 = [H ][CN ] [HCN] Since every dissolved CN - is lso hydrolyzes into HCN up to certin so, [Ag + ]= [CN - ] + [HCN] 6. 0 or, = [CN - [CN ](0.0) ] + 0 [CN ] 6. 0 [CN - ] = (. 0 [CN - ]=.70 - so, [Ag + ]= 6 0 )(6. 0 0.0 6 sp. 0 5 5.9 0 [CN ].7 0 ) Wht is solubility of PbS () ignoring the hydrolysis of ions (b) including the hydrolysis of ions (ssume ph of solution = 7). Given tht: 8 b Pb(OH).5 0 (H S) 0 7 sp(pbs) = 0 7, (HS) = 0 4 M Solution: ) Pbs(S) Pb ++ + S sp = [Pb ++ ] [S ] = S S = S = 7 0 9 S = 8.4 0 5 b) Including hydrolysis: The equilibri of interest re The equilibri of interest re Pb + + H O Pb(OH) + + H O + [Pb(OH)] ; h = [Pb = w ---- (i) b (Pb(OH) 0 4 = 8.5 0 = 6.7 0 7 [H O ] ]
get S + H O HS + OH, HS - - + H O HS + OH; H H [HS ] [OH ] = [S ] w (H S) w = (ii) (H S) = 9. 0 8 [ HS][ OH] = ----- (iii) [HS ] Mss blnce expression re: [Pb + ] o = [Pb + ] + [Pb(OH)] + ------------- () [S -- ] o = [S ] + [HS ] + [H S] ------- (b) Substituting the vlue of [Pb(OH) + ] from eqution (i) into eqution () [Pb ++ ] o = [Pb ++ h [Pb ] ] + [H O ] h = [Pb ++ ] = [H O ] [Pb ] [Pb ++ o ] = -------- (c) h [H O ] Similrly using eqution (ii) nd (iii) in eqution (b) we hve [S ] [S o ] = ---------- (d) h h h [OH] [OH] Now, PbS(s) Pb ++ + S -- sp = [Pb ++ ] [S ] Substituting the vlues of [Pb ++ ] nd [S ] from equtions (c) nd (d), we [Pb ] sp(pbs) o[s ] o = h h h h [HO ] [OH [OH] If y be the mount of PbS dissolve then [Pb + ] o = [S ] o = [ph = 7, [H O + ] = 0 7 = [ OH]] y sp(pbs) = = 7 0 9 7 6.7 0 9.0 7 7 7 0 0 0 On solving, Y =.046 0 0 Problem : Clculte the solubility of MnS in pure wter. Assume hydrolysis of S ions. sp(mns) =.5 0 0 nd of HS re 0 7 nd 0 4 respectively
Solution: Let molr solubility of MnS be XM Mn(s) + Mn S S S 4 S + H O HS + OH W 0 h = 4 0 HS + H O H S + OH W 0 h = 0 7 7 0 As h h 4, first step hydrolysis is lmost complete x = [Mn + ] = [HS ] = [OH ] Consider first step hydrolysis [HS ] [OH ] h [S ] or, [S [HS ] [OH ] ] = = h At equilibrium, [Mn + ] [S ] = sp =.5 0 0 0.5 0 Or n = x x = 6. 0 4 M x Problem : How much solid NSO should be dded to.0 L of wter so tht 0.0005 mole Cd (OH) could just brely dissolve? nd for SO -- complextion with Cd ++ re 8.0 nd.50, respectively. sp (Cd(OH)= 4.50-5 Solution: Cd(OH) Cd ++ + OH - sp = [Cd ++ ] [OH - ] =4.50-5 [Cd(S Cd ++ + S O -- O )] Cd (S O ) = [Cd ][S O Cd (S O ) + S O -- Cd (S O ) -- ; = [Cd(S [Cd][S.50 Assume tht S O -- dose not hydrolyze [Cd ++ ] + [Cd(S O )] + [Cd(S O ) -- ] = 0.00050 [Cd ++ ]+ [Cd ++ ] S O -- ] + [Cd ++ ] [S O -- ] = 0.00050-5 [Cd ++ sp 4.5 0 9 ]= 4.5 0 M [OH ] (0.00) let [S O --- ]=x then, + x+ x 5 0 = 4.5 0 x=0.009 4 9.0 5 O ] O )][S = 8.0 ) ] O ]
wt m.wt.009 wt.009 58.74 Problem : A solution contins mixture of Ag + (0.M) nd Hg (0.M) which re to be seprted by selective precipittion. Clculte the mximum concentrtion of iodide ion t which one of them gets precipitted lmost completely. Wht percentge of tht metl ion is precipitted? sp (AgI) = 8.5 0 7 sp (HgI) =.5 0 6 Solution: Let us first clculte [I ] to precipitte AgI nd Hg I sp[agi] = [Ag + ] [I ] 8.5 0 7 = (0.) [I ] 7 8.5 0 [I ] to precipitte s AgI = 0. sp(hg I ) = [I ] = 8.5 0 6 M [Hg ].5 0 6 = 0. [I ] [I ] to precipitte Hg I = 5.0 0 M [I ] to precipitte AgI is smller. Therefore, Ag I will strt precipitting first. On further ddition of I more AgI will precipitte nd when [I ] 5.0 0 J, Mg I will strt precipitting. The mximum concentrtion of Ag + t this stge will thus be clculted s: sp(agi) = [Ag + ] [I ] 8.5 0 7 = [Ag + ] (5.0 0 ) 7 8.5 0 or, [Ag + ] = 5.0 0 =.7 0 4 M 4.7 0 Percentge of Ag + remined precipitted = 00 = 0.7% 0. Thus percentge of Ag + precipitted = 99.8% Problem 4: Solution: Wht [H + ] must be mintined in sturted HS (0.M) to precipitte CdS but not ZnS, if [Cd + ] = [Zn + ] = 0. (M) initilly? In order to prevent precipitting of ZnS, [Zn + ] [S ] sp (ZnS) = 0 Ionic product or, 0. [S ] 0 or, [S ] 0 0 This is the mximum vlue of [S ] before ZnS will precipitte Let is the mximum vlue of [S ] be x. Thus for H S H + + S [H ][S ] x ( 0 0 ) = = =. 0 [H S] 0.
or, x = [H + ] = 0. (M) No ZnS will precipitte t nd concentrtion of H + greter thn 0.M Problem 5: Solution: Assuming the complete dissocition of HCl nd the led slt, clculte how much HCl is dded to 0.00M led slt solution to just percent precipittion when sturted with HS. The concentrtion of HS in its sturted solution is 0.M (HS) =. 0 sp (PbS) =.4 0 8 We know, sp(pbs) = [Pb + ] [S ] Since led slt is completely dissocited, [Pb + ] is equl to the concentrtion of led slt, i.e. [Pb + ] = 0.00M. If[S ] is the concentrtion of S required to just strt precipittion of PbS. 8.4 0 [S ] = =.4 0 5 0.00 Now the ddition of HCl with suppress the dissocition of H S to tht extent tht [S ] = 4 0 5 (M) HCl is completely ionised, [H + ] = [HCl] Let [HCl] be x. Therefore [H + ] = x H S H + + S At equilibrium [H S] = 0..4 0 5 0. " [H + ] =.4 0 5 + x x [S ] =.4 0 5 [H ] [S ] = [H S]. 0 = x (.4 0 5 5.4 0 x =.80 This ny concentrtion of HCl greter thn.8 M Will just prevent precipittion ) Objective Problem : If the degree of ioniztion of wter be.8 0-9 t 98. Its ioniztion constnt will be (A).8 0-6 (B) 0-4 (C) 0-6 (D).67 0-4 Solution: = (A) w [H O] 4 0 =.8 0 6 55.5
Problem : When solution of benzoic cid ws titrted with NOH the ph of the solution when hlf the cid neutrlized ws 4.. Dissocition constnt of the cid is (A) 6. 0-5 (B). 0 5 (C) 8.7 0 8 (D) 6.4 0 4 Solution: C 6H 5COOH + NOH C 6H 5COON + H O After 0.5 0.5 neutrliztion It is buffer solution of wek cid nd its slt Problem : ph = p + log p = 4. = 6. 0 5 (A) Solution: Old ph = 7 [slt] [cid] 0 - mole of NOH ws dded to 0 litre of wter. The ph will chnge by (A) 4 (B) (C) (D) 7 New [OH ] = 0 New ph = Chnge in ph = 4 (A) 0 = 0 Problem 4: For n queous solution to be neutrl it must hve (A) ph = 7 (B) [H + ]=[OH ] (C) [H + ] = w (D) [H + ] < [OH ] Solution: For neutrl solution [H + ] lwys must be equl to [OH ] (B) Problem 5: If n queous solution t 5 C hs twice s mny OH - s pure wter its poh will be (A) 6.699 (B) 7.07 (C) 7 (D) 6.98 Solution: [OH ] = 0 7 poh = 4 ph or log [OH ] (A) Problem 6: If n cidic indictor HIn Ionises s HIn H + + In. To which mximum ph vlue its solution hs distinct colour chrcteristic of HIn. (A) pin (B) pin ± (C) pin (D) 7
Solution: Problem 7: For colour chrcteristic of H indictor ph = p in log [ HIn ] [In ] Indicter s colour rnge PH = pin (B) Solubility of AgCl in wter, 0.0M CCl, 0.0M NCl nd 0.05M AgNO re S, S, S nd S4 respectively then. (A) S > S > S > S4 (B) S > S > S > S4 (C) S > S = S > S4 (D) S > S > S4 < S Solution: AgCl Ag + + Cl In CCl CCl C + + Cl 0.0 0.0 0.0 In NCl NCl N + + Cl 0.0 0.0 0.0 In AgNO AgNO Ag + + Cl 0.05 0.05 0.05 common ion effect is mximum in AgNO So, S > S > S > S 4 (B) Problem 8: Solution: Problem 9: Wht would be the ph of n mmoni solution if tht of n cetic cid solution of equl strength is.? Assume dissocition constnt for NH & cetic cid re equl. (A). (B) 6.4 (C) 9.6 (D) 0.8 Hint: 4 = p + p b p b = 4. = 0.8 (D) ph of B(OH) solution is. Its solubility product is (A) 0 6 M (B) 4 0 6 M (C) 0.5 0 7 M (D) 5 0 7 M Solution: Since ph = poh = 4 = [OH ] = 0 M We know B (OH) B ++ + OH [B ++ ] = 0 M