Acid Bse Equilirium Review Proof of true understnding of cid se equilirium culmintes in the ility to find ph of ny solution or comintion of solutions. The ility to determine ph of multitude of solutions encompsses mny of the sic knowledge pieces tht mke up the solution equilirium component of Big Ide 6 in the AP Chemistry course description. Also keep in mind tht titrtion curves, which grph the reltionship etween ph nd concentrtions, re creted y series of ph clcultions. Titrtions re very importnt, yet often rushed topic in AP Chemistry, so let s mke sure you see the ig picture. You should e le to find the ph of: 1. A strong cid or se 2. A wek cid or se 3. A slt solution 4. A uffer solution 5. A mixture of strong cid or se with ny of the others (k Invder Prolem) Let s strt y mking sure you cn identify which type of solution is present efore we review the mth. For ech of these solutions, identify which of the solution types (#1-5 ove) re present. 1. NH 3 2. HClO 4 3. HC 2 H 3 O 2 4. NH 4 Cl 5. HNO 3 + NH 3 6. HNO 2 + NNO 2 7. NCHO 2 8. C 2 H 5 NH 2 + C 2 H 5 NH 3 Cl 9. HF 10. KOH + HC 2 H 3 O 2 Now, let s review the mth to clculte the ph for ech type of solution. STRONG ACID OR BASE To find the ph of strong cid or se, tke the log of the cid or se concentrtion. The log of strong cid concentrtion will give the ph. The log of strong se concentrtion will give the poh. The only reson we cn use the concentrtion of the molrity directly is ecuse these cids nd ses re strong. Strong cids nd ses dissocite 100% so for every one mole of cid or se there is one mole of hydronium ions (H + ) nd hydroxide ions (OH - ) respectively. Ex 1) Clculte the ph of 0.450M HCl Ex 2) Clculte the ph of 0.710M KOH Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 1
WEAK ACID OR BASE To find the ph of wek cid or se, write Lowry Brönsted rection y dding the wek cid or wek se to wter. Put tht rection in RICE tle. The given concentrtion is the initil concentrtion of the cid or se rectnt. Solve for the RICE tle x y writing the equilirium expression, plugging in the vlues from the equilirium line of the RICE tle, nd setting it equl to the given K or K. To find the ph/poh, tke the log[x] since x represents the hydronium ion concentrtion in cids nd the hydroxide ion concentrtion in ses. Ex 3) Clculte the ph of 0.150M solution of mmoni. The K of mmoni is 1.80x10-5. Ex 4) Clculte the ph of 0.227M solution of hydrocynic cid (HCN). The K = 6.2 10 10. SALT SOLUTION (HYDROLYSIS) To find the ph of slt solution, first determine which ion within the slt will undergo hydrolysis. In other words, which ion from the slt will rect with wter to form compound tht won t significntly dissocite? A quick wy to figure tht out look t ech ion in the slt nd sk yourself, if this is pired with n H + or OH - will tht sustnce e strong or wek? We re looking for the one tht mkes wek product ecuse tht product will not re-ionize nd will thus chnge the H + /OH - concentrtion in the solution. Once you ve identified the wek ion, write n eqution for tht ion splitting wter molecule nd use tht s the Rection line in RICE tle. Proceed with solving for ph s you do with ny other wek cid or se prolem. One other importnt component of hydrolysis prolems is the conversion of K to K or vice vers. The question prompts given on the AP exm will provide the K or K of the conjugte cid or se, not the K of the ion you re using in the hydrolysis eqution. The K or K of the hydrolyzing ion needs to e determined using K w = K x K. This is necessry step for every hydrolysis prolem! Ex 5) Clculte the ph of 0.95M KC 3 H 5 O 3 solution (potssium lctte). The K, for lctic cid, HC 3 H 5 O 3 is 7.1x10-12 Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 2
BUFFER SOLUTION Method 1: If solution contins wek cid nd its conjugte se, or wek se nd its conjugte cid, then uffer is present. Using the Henderson-Hsselch (HH) eqution is one wy to determine ph of uffer solution. There is ph version nd poh version of the eqution. The cid version of this eqution is provided to students on the AP Eqution insert of the AP Chemistry exm. You re llowed ccess to tht formul sheet for the durtion of the test. You cn lso progrm this eqution into your clcultor! A ph pk log [ ] HB = + poh pk log [ + = + ] [ HA] [ B] Using two versions of the HH eqution prevents you from getting the concentrtions of the cids/ses nd their conjugtes confused when plugging in to the log rtio. Alwys put the ION concentrtion in the numertor nd the cid/se concentrtion in the denomintor. Sy ION ON TOP! over nd over to help you rememer. ion ph pk log [ ] ion = + poh = pk + log [ ] [ cid] [ se] Ex 6) Clculte the ph of solution which is 0.53M in HC 6 H 4 NO 2 nd 0.50M NC 6 H 4 NO 2. The K for nicotinic cid is 1.7x10-5. Ex 7) Clculte the ph of solution which is 0.245M in NH 3 nd 0.245M in NH 4 Cl. The K for mmoni is 1.8x10-5. Method 2: Another method for solving uffer prolems using n eqution similr to HH ut without the logrithms. (Actully, when deriving the HH eqution, this eqution occurs just one step prior to tking the logrithms tht generted the HH eqution). In this pproch we use modified K expression tht isoltes the [H + ], rther thn the K : H K cid + [ ] [ ] = [ se] So, in your uffer system, strt y identifying the component tht is esily recognizle s either wek cid or se. Tht concentrtion is the one tht will go in the [cid] or [se] rckets. Then, the conjugte of tht wek cid/se is wht will go in the remining rcket. Alwys plug in the K nd you will lwys e solving for the [H + ]. If they give you K, just convert it to K with K w = K x K nd then plug in the K. There is no need to consider sic version of this eqution ecuse we re lmost lwys interested in the [H + ] even if it s sic solution. Once you hve the [H + ], convert it to ph with the stndrd log[h + ] nd you re done. Ex 6) Clculte the ph of solution which is 0.53M in HC 6 H 4 NO 2 nd 0.50M NC 6 H 4 NO 2. The K for nicotinic cid is 1.7x10-5. Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 3
Ex 7) Clculte the ph of solution which is 0.245M in NH 3 nd 0.245M in NH 4 Cl. The K for mmoni is 1.8x10-5. MIXTURE OF STRONG ACID/BASE WITH ANYTHING ELSE Tret the ddition of strong cid or se to system lredy t equilirium s n invder. This invsion cretes wr. The wr is represented y the one-wy ( ) stoichiometric rection etween the invder nd the system t equilirium. Once the wr is over, the stoichiometric survivors remining will then determine the ph of the solution. This is multistep process tht will e shown in the three exmples provided for you elow, with ech exmple hving different outcome. Ex 8) 125mL of 0.525M HCl is mixed with 125mL of uffer system contining 1M cetic cid nd 1M sodium cette. Find the ph of the solution. The K of cetic cid is 1.8x10-5. Before you strt, since this is uffer, mke sure you cn identify the components: The cid is nd the se is Step 1: Write the invsion eqution s the strong cid enters the uffer system. (Think: which component of the uffer will neutrlize this invder? And note, this is NOT n equilirium rection rection only) Step 2: Figure out the stoichiometric survivors using quick mole stoichiometry tle such s: Initil moles Chnge in moles Finl moles Step 3: Bsed on the survivors, mke pln. Are there oth [cid] nd [se] components of uffer? Or, is there just one survivor? If it s still uffer, plug the finl moles of ech piece into either HH or uffer eqution nd solve for ph. If there is only one survivor left, find its ph ppropritely s n cid, se or slt. Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 4
Ex 9) 125mL of 0.500M NOH is mixed with 125mL of 0.500M cetic cid. Find the ph of the solution. The K of cetic cid is 1.8x10-5. Step 1: Write the invsion eqution s the strong cid enters the cid system. (Note, this is NOT n equilirium rection rection only) Step 2: Figure out the stoichiometric survivors using quick mole stoichiometry tle such s: Initil moles Chnge in moles Finl moles Step 3: Bsed on the survivors, mke pln. Are there oth [cid] nd [se] components of uffer? Or, is there just one survivor? If it s still uffer, plug the finl moles of ech piece into either HH or uffer eqution nd solve for ph. If there is only one survivor left, find its ph ppropritely s n cid, se or slt. Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 5
Ex 10) 200mL of 0.500M NOH is mixed with 125mL of 0.500M cetic cid. Find the ph of the solution. The K of cetic cid is 1.8x10-5. Step 1: Write the invsion eqution s the strong cid enters the cid system. (Note, this is NOT n equilirium rection rection only) Step 2: Figure out the stoichiometric survivors using quick mole stoichiometry tle such s: Initil moles Chnge in moles Finl moles Step 3: Bsed on the survivors, mke pln. Are there oth [cid] nd [se] components of uffer? Or, is there just one survivor? If it s still uffer, plug the finl moles of ech piece into either HH or uffer eqution nd solve for ph. If there is only one survivor left, find its ph ppropritely s n cid, se or slt. Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 6
Acid Bse Equilirium Review Proof of true understnding of cid se equilirium culmintes in the ility to find ph of ny solution or comintion of solutions. The ility to determine ph of multitude of solutions encompsses mny of the sic knowledge pieces tht mke up the solution equilirium component of Big Ide 6 in the AP Chemistry course description. Also keep in mind tht titrtion curves, which grph the reltionship etween ph nd concentrtions, re creted y series of ph clcultions. Titrtions re very importnt, yet often rushed topic in AP Chemistry, so let s mke sure you see the ig picture. You should e le to find the ph of: 1. A strong cid or se 2. A wek cid or se 3. A slt solution 4. A uffer solution 5. A mixture of strong cid or se with ny of the others (k Invder Prolem) Let s strt y mking sure you cn identify which type of solution is present efore we review the mth. For ech of these solutions, identify which of the solution types (#1-5 ove) re present. 1. NH 3 wek se 2. HClO 4 strong cid 3. HC 2 H 3 O 2 wek cid 4. NH 4 Cl slt 5. HNO 3 + NH 3 mixture of strong cid nd wek se 6. HNO 2 + NNO 2 uffer 7. NCHO 2 slt 8. C 2 H 5 NH 2 + C 2 H 5 NH 3 Cl uffer 9. HF wek cid 10. KOH + HC 2 H 3 O 2 mixture of strong se nd wek cid Now, let s review the mth to clculte the ph for ech type of solution. STRONG ACID OR BASE To find the ph of strong cid or se, tke the log of the cid or se concentrtion. The log of strong cid concentrtion will give the ph. The log of strong se concentrtion will give the poh. The only reson we cn use the concentrtion of the molrity directly is ecuse these cids nd ses re strong. Strong cids nd ses dissocite 100% so for every one mole of cid or se there is one mole of hydronium ions (H + ) nd hydroxide ions (OH ) respectively. Ex 1) Clculte the ph of 0.450M HCl Since this is strong cid nd dissocites 100% ssume tht the concentrtion of the cid equls the concentrtion of the hydronium ion in solution. ph= log[h + ] ph= log[0.450] ph= 0.347 Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 1
Ex 2) Clculte the ph of 0.710M KOH Since this strong se nd dissocites 100% ssume tht the concentrtion of the cid equls the concentrtion of the hydroxide ions in solution poh= log[oh ] poh= log[0.710] poh= 0.347 ph = 14 0.347 ph= 13.851 WEAK ACID OR BASE To find the ph of wek cid or se, write Lowry Brönsted rection y dding the wek cid or wek se to wter. Put tht rection in RICE tle. The given concentrtion is the initil concentrtion of the cid or se rectnt. Solve for the RICE tle x y writing the equilirium expression, plugging in the vlues from the equilirium line of the RICE tle, nd setting it equl to the given K or K. To find the ph/poh, tke the log[x] since x represents the hydronium ion concentrtion in cids nd the hydroxide ion concentrtion in ses. Ex 3) Clculte the ph of 0.150M solution of mmoni. The K of mmoni is 1.80x10 5. + - [NH ][OH ] 4 K= [NH ] 3 The use of RICE tle will led to the elow sustitutions into the equilirium expression. The x in the denomintor is neglected due to mthemticl insignificnce. 2 x K = where x = [OH ] M x 2 x 1.80 10 = 0.150 3 x = 1.64 10 = [OH ] 3 poh = log(1.64 10 ) = 2.784 ph = 11.216 Ex 4) Clculte the ph of 0.227M solution of hydrocynic cid (HCN). The K = 6.2 10 10 The use of RICE tle will led to the elow sustitutions into the equilirium expression. The x in the denomintor is neglected due to mthemticl insignificnce. 2 x + K = where x = [H ] M x 2 10 x 6.20 10 = 0.227 x = 1.19 10 = [H + ] ph = log(1.19 10 ) = 4.926 Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 2
SALT SOLUTION (HYDROLYSIS) To find the ph of slt solution, first determine which ion within the slt will undergo hydrolysis. In other words, which ion from the slt will rect with wter to form compound tht won t significntly dissocite? A quick wy to figure tht out look t ech ion in the slt nd sk yourself, if this is pired with n H + or OH will tht sustnce e strong or wek? We re looking for the one tht mkes wek product ecuse tht product will not re-ionize nd will thus chnge the H + /OH concentrtion in the solution. Once you ve identified the wek ion, write n eqution for tht ion splitting wter molecule nd use tht s the Rection line in RICE tle. Proceed with solving for ph s you do with ny other wek cid or se prolem. One other importnt component of hydrolysis prolems is the conversion of K to K or vice vers. The question prompts given on the AP exm will provide the K or K of the conjugte cid or se, not the K of the ion you re using in the hydrolysis eqution. The K or K of the hydrolyzing ion needs to e determined using K w = K x K. This is necessry step for every hydrolysis prolem! Ex 5) Clculte the ph of 0.95M KC 3 H 5 O 3 solution (potssium lctte). The K, for lctic cid, HC 3 H 5 O 3 is 7.1x10 12 The dissocited ion from the slt tht undergoes hydrolysis is C 3 H 5 O 3. Add tht ion to wter nd put it in RICE tle long with the following eqution. C 3 H 5 O 3 + H 2 O HC 3 H 5 O 3 + OH K is needed in the eqution insted of K. 14 K w 1 10 3 K = = = 1.4 10 12 K 7.1 10 Equilirium expression from rection: [HC H O ][OH ] 3 5 3 K = [C H O ] 3 5 3 The use of RICE tle will led to the elow sustitutions into the equilirium expression. 2 x 3 K = = 1.4 10 0.95 2 x = [OH ] = 3.6 10 2 poh = log(3.6 10 ) = 1.47 ph = 14 1.47 = 12.53 Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 3
BUFFER SOLUTION Method 1: If solution contins wek cid nd its conjugte se, or wek se nd its conjugte cid, then uffer is present. Using the Henderson-Hsselch (HH) eqution is one wy to determine ph of uffer solution. There is ph version nd poh version of the eqution. The cid version of this eqution is provided to students on the AP Eqution insert of the AP Chemistry exm. You re llowed ccess to tht formul sheet for the durtion of the test. You cn lso progrm this eqution into your clcultor! A ph pk log [ ] HB = + poh pk log [ + = + ] [ HA] [ B] Using two versions of the HH eqution prevents you from getting the concentrtions of the cids/ses nd their conjugtes confused when plugging in to the log rtio. Alwys put the ION concentrtion in the numertor nd the cid/se concentrtion in the denomintor. Sy ION ON TOP! over nd over to help you rememer. ion ph pk log [ ] ion = + poh = pk + log [ ] [ cid] [ se] Ex 6) Clculte the ph of solution which is 0.53M in HC 6 H 4 NO 2 nd 0.50M NC 6 H 4 NO 2. The K for nicotinic cid is 1.7x10 5. Solution: Since this is solution which contins wek cid nd its conjugte se we will plug numers in the cidic version of Henderson Hsselch (HH) nd solve. ph = pk + log [ion] [cid] ph log(1.7 10 ) log [0.50] = + = 4.74 [0.53] Ex 7) Clculte the ph of solution which is 0.245M in NH 3 nd 0.245M in NH 4 Cl. The K for mmoni is 1.8x10 5. Solution: Since this is solution tht contins wek cid nd it s conjugte se we will plug numers into the sic version of HH nd solve. poh = pk + log [ion] [se] poh log(1.8 10 ) log [0.245] = + = 4.74 [0.245] ph = 14 4.74 = 9.26 You cn see ove tht the rtio of se to conjugte cid is 1. When this is the cse poh=pk. This is very importnt concept in titrtions ecuse it s considered to e the titrtion s midpoint or hlfwy point. Method 2: Another method for solving uffer prolems using n eqution similr to HH ut without the logrithms. (Actully, when deriving the HH eqution, this eqution occurs just one step prior to tking the logrithms tht generted the HH eqution). In this pproch we use modified K expression tht isoltes the [H + ], rther thn the K : H K cid + [ ] [ ] = [ se] Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 4
So, in your uffer system, strt y identifying the component tht is esily recognizle s either wek cid or se. Tht concentrtion is the one tht will go in the [cid] or [se] rckets. Then, the conjugte of tht wek cid/se is wht will go in the remining rcket. Alwys plug in the K nd you will lwys e solving for the [H + ]. If they give you K, just convert it to K with K w = K x K nd then plug in the K. There is no need to consider sic version of this eqution ecuse we re lmost lwys interested in the [H + ] even if it s sic solution. Once you hve the [H + ], convert it to ph with the stndrd log[h + ] nd you re done. Ex 6) Clculte the ph of solution which is 0.53M in HC 6 H 4 NO 2 nd 0.50M NC 6 H 4 NO 2. The K for nicotinic cid is 1.7x10 5. In this solution the HC 6 H 4 NO 2 is recognizle [cid] nd thus the NC 6 H 4 NO 2 must e the [se]. M [H ] 1.7 10 [0.53 ] + = = 1.8 10 [0.50 M] ph = log(1.8 10 ) = 4.74 Ex 7) Clculte the ph of solution which is 0.245M in NH 3 nd 0.245M in NH 4 Cl. The K for mmoni is 1.8x10 5. In this solution the NH 3 is recognizle [se] nd thus the NH 4 Cl must e the [cid]. Since we will lwys use the K in this eqution, you ll need to convert K to K using K w =K xk. Sve yourself step nd just use the K w /K frction in the K spot of the formul. 14 1 10 + [0.245 M] 10 [H ] = 5.6 10 5 1.8 10 = [0.245 M] 10 ph = log(5.6 10 ) = 9.25 And, did you notice wht hppened when the [cid]/[se] rtio is equl to 1? Yep, the [H + ]=K, or if we tke the log of oth, the ph=pk. This represents the est uffer ecuse it contins equivlent numers of cid nd se prticles tht could del with n invder from either side. Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 5
MIXTURE OF STRONG ACID/BASE WITH ANYTHING ELSE Tret the ddition of strong cid or se to system lredy t equilirium s n invder. This invsion cretes wr. The wr is represented y the one-wy ( ) stoichiometric rection etween the invder nd the system t equilirium. Once the wr is over, the stoichiometric survivors remining will then determine the ph of the solution. This is multistep process tht will e shown in the three exmples provided for you elow, with ech exmple hving different outcome. Ex 8) 125mL of 0.525M HCl is mixed with 125mL of uffer system contining 1M cetic cid nd 1M sodium cette. Find the ph of the solution. The K of cetic cid is 1.8x10 5. Before you strt, since this is uffer, mke sure you cn identify the components: The cid iscetic cid nd the se is cette ion Step 1: Write the invsion eqution s the strong cid enters the uffer system. (Think: which component of the uffer will neutrlize this invder? And note, this is NOT n equilirium rection rection only) The first step is to write rection representing the invsion of the strong cid into the uffer system. The strong cid H + or the strong se OH will lwys e rectnt. In this cse the invder H + will rect with only one component of the uffer ut will produce the other. This is NOT n equilirium rection. This rection will go to completion. After the invsion the wr will eventully come to n end. Writing these first equtions is usully the sticking point for most students. RXN: H + + C 2 H 3 O 2 HC 2 H 3 O 2 Step 2: Figure out the stoichiometric survivors using quick mole stoichiometry tle such s: The next step is to mke stoichiometry tle which somewht resemles RICE tle ut insted of concentrtions or pressures, it s comprison of moles. From the tle it cn e determined how mny moles of ech rectnt rect nd how much of wht is left over. H + C 2 H 3 O 2 HC 2 H 3 O 2 Initil moles 0.0656 0.125 0.125 Chnge in moles 0.0656 0.0656 +0.0656 Finl moles 0 0.0594 0.1906 Step 3: Bsed on the survivors, mke pln. Are there oth [cid] nd [se] components of uffer? Or, is there just one survivor? If it s still uffer, plug the finl moles of ech piece into either HH or uffer eqution nd solve for ph. If there is only one survivor left, find its ph ppropritely s n cid, se or slt. Now tht the wr is over, study wht s remining. The invder lost. There is none left. Since there is still moles of cid nd conjugte se left over in the solution we now solve this s n ordinry uffer prolem. The only thing to wtch out for is tht the tle ws clculted in MOLES. We need to convert these moles to concentrtion y dividing y the totl finl volume in the mjority of cses. However, since oth of our pproches rely on rtio of the two components, dividing y the totl volume will just cncel out nywy, so we cn sve step nd just plug in the moles rther thn convert to molrity. Method 1: Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 6
ph = pk + log [ion] [cid] ph log(1.8 10 ) log [0.0594] = + = 4.23 [0.1906] Method 2: [0.1906] [H ] = ( 1.8 10 ) = 5.8 10 [0.0594] + ph = log(5.8 10 ) = 4.23 Ex 9) 125mL of 0.500M NOH is mixed with 125mL of 0.500M cetic cid. Find the ph of the solution. The K of cetic cid is 1.8x10 5. Step 1: Write the invsion eqution s the strong cid enters the cid system. (Note, this is NOT n equilirium rection rection only) The first step is to write rection representing the invsion of the strong se into wek cid. In this cse the invder OH will rect with the only thing there is to rect with. This is n esier eqution to write thn the lst. RXN: OH + HC 2 H 3 O 2 C 2 H 3 O 2 + H 2 O Step 2: Figure out the stoichiometric survivors using quick mole stoichiometry tle such s: OH HC 2 H 3 O 2 C 2 H 3 O 2 Initil moles 0.0625 0.0625 0 Chnge in moles 0.0625 0.0625 +0.0625 Finl moles 0 0 0.0625 Step 3: Bsed on the survivors, mke pln. Are there oth [cid] nd [se] components of uffer? Or, is there just one survivor? If it s still uffer, plug the finl moles of ech piece into either HH or uffer eqution nd solve for ph. If there is only one survivor left, find its ph ppropritely s n cid, se or slt. In this scenrio the only thing left over is n ion tht undergoes slt hydrolysis. Which mens we re going to tret this just like finding the ph of slt solution. First, tke the moles remining nd convert it to molrity so tht it looks like the other slt solution prolems you ve worked. 0.0625mol/0.250L = 0.250M C 2 H 3 O 2 Rememer, this is just regulr slt hydrolysis prolem. (Don t forget to convert K to K ) 14 K w 1 10 10 K = = = 5.56 10 K 1.8 10 The dissocited ion from the slt tht undergoes hydrolysis is C 2 H 3 O 2. Add tht ion to wter nd this is wht hppens: C 2 H 3 O 2 + H 2 O HC 2 H 3 O 2 + OH This is the R eqution for your RICE tle, if you need it. And, it mkes the equilirium expression: [HC H O ][OH ] 2 3 2 K = [C H O ] 2 3 2 Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 7
From rice tle we sustitute to get (neglecting x): 2 10 x 5.56 10 = 0.250 x = [ OH ] = 1.18 10 poh = log(1.18 10 ) = 4.92 ph = 14 4.92 = 9.08 Note: This is significnt scenrio in n cid se titrtion ecuse this represents the equivlence point of titrtion nd shows why the ph in wek/strong sitution will not hve n equivlence point of 7. Ex 10) 200mL of 0.500M NOH is mixed with 125mL of 0.500M cetic cid. Find the ph of the solution. The K of cetic cid is 1.8x10 5. Step 1: Write the invsion eqution s the strong cid enters the cid system. (Note, this is NOT n equilirium rection rection only) The first step is to write rection representing the invsion of the strong se into wek cid. In this cse the invder OH will rect with the only thing there is to rect with. This is n esier eqution to write thn the lst. RXN: OH + HC 2 H 3 O 2 C 2 H 3 O 2 + H 2 O Step 2: Figure out the stoichiometric survivors using quick mole stoichiometry tle such s: OH HC 2 H 3 O 2 C 2 H 3 O 2 Initil moles 0.100 0.0625 0 Chnge in moles 0.0625 0.0625 +0.0625 Finl moles 0.0375 0 0.0625 Step 3: Bsed on the survivors, mke pln. Are there oth [cid] nd [se] components of uffer? Or, is there just one survivor? If it s still uffer, plug the finl moles of ech piece into either HH or uffer eqution nd solve for ph. If there is only one survivor left, find its ph ppropritely s n cid, se or slt. In this scenrio there is some strong se left over. The invder wins!! Any time there is strong se or strong cid left over, you just use those left over moles to find ph. Any other moles of ny other sustnce will not significntly chnge the ph if there re strong ions present. DON T FORGET to convert the moles to concentrtion y dividing y totl volume efore finding ph. 0.0375mol/0.325L = 0.115M OH poh= log[oh ] poh= log[0.115] poh= 0.94 ph = 14 0.94 ph= 13.06 Copyright 2016 Ntionl Mth + Inititive, Dlls, Texs. All rights reserved. Visit us online t www.nms.org 8