Solution:
Calculation of pH for HV:
HV + H2O = V- + H3O+
0.288 ---------0 ---------0. (initial)
0.288 - X ---- X ------ X (final)
Ka1 = X . X / (0.288 - X)
Since, acid is weak, hence X is neglected from denominator)
Ka1 = X^2./ 0.288
X^2 = 0.288 x Ka1 = 0.288 x 5.18 x 10^-3
X^2 = 1.492 x 10^-3 = 0.1492 x 10^-4
X = 0.386 x 10^-2 = 3.86 x 10^-3
Therefore,
[H3O+] = [H+] = 3.86 x 10^-3 M
Thus,
pH = - log [H+] = - log 3.86 x 10^-3
pH = 3 - log 3.86 = 3 - 0.59 = 2.41
pH = 2.41
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