The ionization constant of propanoic acid is 1.32×10−5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl ?
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Solution
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Let x be the degree of dissociation. x=√Kac=√1.32×10−50.05=0.0162 [H+]=cx=0.05×0.0162=8.12×10−4 pH=log[H+]=−log(8.12×10−4)=3.09 If the solution is 0.01 M in HCl also then, x=√Kac=√1.32×10−50.01=0.00132
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