Consider the titration of 25.0 mL of a 0.130 M hydrocyanic acid (HCN) solution w

Question

Consider the titration of 25.0 mL of a 0.130 M hydrocyanic acid (HCN) solution with a 0.100 M potassium hydroxide solution. Calculate the pH (a) before the titration, (b) at the half-equivalence point, (c) at the equivalence point. (Hint: The Ka for HCN is 4.9

Explanation / Answer

a) HCN ---> H+ CN- , [HCN] = (0.13-x) , [H+]=CN-] = x , Ka = {H+][CN-]/[HCN] , 4.9 x10^ 10 = x^2/( 0.13-x) , x = 7.98 x10^ -6 = [H+] , pH = -log( 7.98 x10^ -6) = 5.098, b) at half equivalence point pH = pka = -log( 4.9 x10^ -10) = 9.3, c) at equialence point all acid molesules will get neutralised by base , so we are left with only salt solution i.e sodium cynide aq solution, Na+ CN- + H2O ---> OH- + HCN + Na, [OH-]=[HCN] = x , [NaCN] = 0.2-x Kb = Kw/Ka = 10^ -14 /4.9 x10^ -10 = 2.04 x10^ -5 , where [NaCN] = 2 x .1 = 0.2 M ( since both acid an base add in equal proportions) , Kb = ( x^2)/( 0.2-x) = 2.04 x10^ -5 , x = 0.002 = [OH-] , pOh = -log( 0.002) = 2.7 , pH = 14-2.7 =11.3

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