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Question

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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.160 M HCIO(ag) with 0.160 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 35.0 mL of KoH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOH

Explanation / Answer

mol of M^2+ = 0.170 mole
mol of CN = 1L x 1.44mol/L = 1.44 mol

. . . . . . . . .m^2+ . . .+ . . . .6 cn <===> m(cn)6
initial : , , , ,0.170 . . . . . . . . .1.44
react . . . . . . n . . . . . . . . . . . 6n
equilibrium : 0.170-n . . . . . .1.44 - 6n . . . . . . n

K = [ m(CN)6 ] / ( [M^2+][CN]^6 )
2.5

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