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Given:C=0.45M We know that the dissociation constant (Ka) of HCN=6.28 X10-10 Solution: KCN is salt of strong acid and weak base. Kh=Kw/Ka Kh=1 X 10-14/6.2 * 10-10 Kh =1.613 X10-5 ---- Kh=h2C h2=Kh/C =1.613 X 10-5/0.45 =0.3584 X10-4 ---- h=\/0.3584 X 10-4 h =0.6 X 10-2 ---- For a salt of weak acid and strong base, [OH-] = h X C =0.6 * 10-2 X 0.45 [OH-] =2.7 X 10-3 ---- pOH = -log[OH-] =-[log (2.7 X 10-3)] =-[-3 log 10 + log 2.7] =-[-3 + 0.4314] =3 - 0.4314 pOH = 2.5686 ---- pH + pOH = 14 pH = 14 - pOH =14 - 2.5686 pH = 11.4314 ---- Therefore the pH of 0.45M KCN is = 11.4314.

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