temperature is known as ionic product of water. It is designated as K w. Value of K w

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Ionic product of ter The product of concentrtions of H nd OH ions in ter t prticulr temperture is knon s ionic product of ter. It is designted s K. H O H 1 OH ; H 57.3 kjm The vlue of [H ][OH ] K ; K[HO] [H ][OH ]; K [H ][OH ] [H O] K increses ith the increse of temperture, i.e., the concentrtion H + nd OH ions increses ith increse in temperture. The vlue of Temperture ( o C) Vlue of K 0 0.11 0.31 5 1.00 0 7.50 K t 1 o 5 C is. Since pure ter is neutrl in nture, H ion concentrtion must e equl to OH ion concentrtion. 7 [H ] [OH ] x or [H ][OH ] x 1 or x 1 M or [H ] [OH ] 1 mole litre 7 1 o 7 This shos tht t 5 C, in 1 litre only mole of ter is in ionic form out of totl of pproximtely 55.5 moles. Thus hen, [H ] [OH ]; the solution is neutrl [H ] [OH ]; the solution is cidic

[H ] [OH ]; the solution is sic Hydrogen ion concentrtion ph scle Sorensen, Dnish iochemist developed scle to mesure the cidity in terms of concentrtions of H in solution. As defined y him, ph of solution is the negtive logrithm to the se of the concentrtion of H + ions hich it contins. ph log[h ] or 1 ph log [H ] A ph scle rnges from 0 to [H3O + 0 1 3 4 5 6 7 8 9 11 1 13 ACIDIC CHARACTER INCREASES BASIC CHARACTER 0 ph 7 1 3 4 5 NEUTRAL 6 8 9 11 1 13 Just s ph indictes the hydrogen ion concentrtion, the poh represents the hydroxyl ion concentrtion, i.e., poh log[oh ] Considering the reltionship, [H ][OH ] K 1 Tking log on oth sides, e hve log[h ] log[oh ] log K log(1 ) or log[h ] log[oh ] log K log(1 )

or ph poh pk i.e., sum of ph nd poh is equl to in ny queous solution t o 5 C. The ove discussion cn e summrised in the folloing mnner, [H + ] [OH ] ph poh Acidic solution Neutrl solution Bsic solution > 7 < 7 < 7 > 7 7 7 7 7 < 7 > 7 > 7 < 7 ph of some mterils Mteril ph Mteril ph Gstric juice 1.4 Rin ter 6.5 Lemon juice.1 Pure ter 7.0 Vinegr.9 Humn sliv 7.0 Soft drinks 3.0 Blood plsm 7.4 Beer 4.5 Ters 7.4 Blck coffee 5.0 Egg 7.8 Co s milk 6.5 Household mmoni 11.9

Limittions of ph scle (i) ph vlues of the solutions do not give us immedite ide of the reltive strengths of the solutions. A solution of ph =1 hs hydrogen ion concentrtion 0 times tht of solution ph = 3 (not three times). A 5 4 N HCl 5 is tice concentrted of N HCl solution, ut the ph vlues of these solutions re 4.40 nd 4.70 (not doule). (ii) ph vlue zero is otined in 1N solution of strong cid. If the concentrtion is N, 3N, N, etc. the respective ph vlues ill e negtive. 8 (iii) A solution of n cid hving very lo concentrtion, sy N, cn not hve ph 8, s shon y ph formul ut the ctul ph vlue ill e less thn 7. pk vlue : p stnds for negtive logrithm. Just s H nd OH ion concentrtions rnge over mny negtive poers of, it is convenient to express them s ph or poh, the dissocition constnt (K) vlues lso rnge over mny negtive poers of nd it is convenient to rite them s pk. Thus, pk is the negtive logrithm of dissocition constnt. pk pk log K log K nd Wek cids hve higher pk vlues pk vlues. Similrly ek ses hve higher For ny conjugte cid-se pir in queous solution, K K K pk pk pk (t 98 K)

8 Clcultion the ph of M HCl If e use the reltion ph log[h3o ] e get ph equl to 8, ut this is not correct ecuse n cidic solution connot hve ph greter thn 7. In this condition H concentrtion of ter cnnot e neglected. Therefore, [H ] H H totl Acid ter Since HCl is strong cid nd completely ionised, [H ] 1 HCl 8 [H ] HO 7 8 7 [H ] [H ] [H ] 8 8 [1 ] 11 totl HCl HO 8 ph log log11 6.958 8 Similrly if NOH concentrtion is M Then, [OH ] [ ] [ ] 8 7 totl NOH HO ; poh 6.96 ph 7.04 8 [OH ] 11 Note : Solutions my hve negtive ph [C M, ph 1] H nd even ph ove [C M, ph 15] OH, ut in such concentrted solutions the ctivity coefficients should e used. Hence the ph rnge t 98K is tken s 0 to for most of the prcticl purposes : Acidity of n queous solution 1/pH ph of pure ter decreses ith increse of temperture e.g., ph of oiling ter is 6.56. ph of the solution of ek cid or ek se depends upon its degree of ioniztion (). The reltionship my e derived s follos tking the exmple of ek cid HA.

HA H A Initil conc. C Conc. t eq. C(1 ) C C [H ] C ph log[h ] logc But K C or C K / ph log(k / ) If pk (or pk ) vlue is negtive, it indictes tht the cid (or se) is completely ionised i.e., cids nd ses re strong. Exmples sed on ph, poh, pk, pk nd pk Exmple 1 : 5.5 mol litre 3 1 If the hydrogen ion concentrtion of given solution is, the ph of the solution ill e ().6 () 3.40 (c) 3.75 (d).76 3 Solution: () [H ] 5.5 mole/ litre ph log[h ] ; ph.6 3 ph log[5.5 ] Exmple : The ph of 0.1M solution of ek monoprotic cid 1% ionized is () 1 () (c) 3 (d) 4

Solution: (c) C 0.1M ; 1% 1 [H ] C 0.1 0 3 ph log[h ] or ph log[ ]; ph 3 3 Exmple 3 : The ph of solution is 5.0. To this solution sufficient cid is dded to decrese the ph to.0. The increse in hydrogen ion concentrtion is () 0 times () 00 times (c).5 times (d) times 5 Solution: () [H ] ;[H ] Initil finl Increse in [H ] = 00 times 5 Exmple 4 : ph vlue of N / NOH solution is () () 11 (c) 1 (d) 13 Solution: (d) In N NOH 1 hve [OH ] M 1 poh log[oh ] or poh log[ ] poh 1 nd then, ph poh or ph poh 13 8 Exmple 5 : Wht ill e the ph of M HCl solution () 8.0 () 7.0 (c) 6.98 (d).0

Solution: (c) HO [H ] [OH ]; HCl [H ] [Cl ] Totl 7 1 [H ] [H ] [H ] [1 ] 7 8 HO HCl 7 11 [H ] 7 11 ph log[h ] log ; ph 6.98 Exmple 6 : By dding 0ml 0.1N HCl to 0ml 0.001N KOH. The ph of the resulting solution ill e () () 1.3 (c) 0 (d) 7 Solution: () 0ml of 0.1 3 0.1N HCl 0g eq. g eq. 00 0ml of 0.001 5 0.001N KOH 0 gm eq. g eq. 00 HCl left unneutrlised 3 5 ( 3 3 3 ) (1 0.01) 0.99 1.98 g eq. Volume of solution 40 ml. 3 1.98 [HCl] 00M 4.95 40 ph log4.95 0.7 1.3 Exmple 7 : For ek cid HA ith dissocition constnt 9, poh of its 0.1M solution is () 9 () 3 (c) 11 (d)

9 Solution: () Dissocition constnt of HA HA H A [H 9 1 ] K C 5 [H ] 5 ph 5 ph poh poh ph 5 ; poh 9 Exmple 8 : If pk for fluoride ion t o 5 C is.83, the ionistion constnt of hydrofluoric cid in ter t this temperture is 3 () 1.74 3 () 3.5 4 (c) 6.75 (d) 5.38 Solution: (c) K K K We kno tht pk log K or.83 log K 11 K Antilog (.83) 1.48 K K K 1.48 11 6.75 4 Exmple 9 : The vlue of K is 9.55 t certin temperture. Clculte the ph of ter t this temperture. () 6.51 () 9.73 (c).8 (d) 4.96

Solution: () Here e re given, K 9.55 No, s for ter, [H ] [OH ] K [H ] [OH ] [H ].[H ] [H ] i.e., [H 7 ] 9.55 or [H ] 9.55 3.09 M 7 ph log[h ] log (3.09 ) 6.51 7 [log3.09 log ] [0.49 7] Exmple : If pk = 13.36 t 50 o C, the ph of ter t the sme temperture is () 7.00 () 6.68 (c) 7.63 (d) 6.00 Solution: () ph poh pk For HO, ph poh x x 13.36 ; x 13.36 13.36 x 6.68 Exmple 11 : Ho mny grms of NOH must e dissolved in one litre of the solution to give it ph vlue of 1 () 0.8 g () 1.8 g (c) 0.4 g (d) 0.6 g

Solution: (c) ph log[h ] log[h ] ph 1 [H 1 ] ntilog ( 1) g ions/litre But e kno tht, [H ][OH ] K K [OH ] g ions/litre 1 [H ] Since NOH is strong electrolyte, it undergoes complete ioniztion s, NOH N OH [OH ] [NOH] M Mol. t. of NOH 40 Amount of NOH dissolved per litre 40 0.4 g