Question

It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionisation constant of the acid and its pK_a.

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Solution

The dissociation of organic acid may be represented as,

HA space rightwards harpoon over leftwards harpoon space space straight H to the power of plus space plus space straight A to the power of minus pH space equals space minus log space open square brackets straight H to the power of plus close square brackets or space space log space open square brackets straight H to the power of plus close square brackets space equals space minus pH space equals space minus 4.15 space equals space 5 with bar on top. space 85 therefore space space space space open square brackets straight H to the power of plus close square brackets space equals space antilog space left parenthesis 5 with bar on top.85 right parenthesis space equals space 7.09 space cross times space 10 to the power of negative 5 end exponent straight M space space space space space space space space space space space space space space space equals space 7.08 space cross times space 10 to the power of negative 5 end exponent straight M

straight K subscript straight a space equals space fraction numerator open square brackets straight H to the power of plus close square brackets space open square brackets straight A to the power of minus close square brackets over denominator open square brackets HA close square brackets end fraction space space space space space space equals fraction numerator left parenthesis 7.08 space cross times 10 to the power of negative 5 end exponent right parenthesis thin space left parenthesis 7.08 space cross times space 10 to the power of negative 5 end exponent right parenthesis over denominator 10 to the power of negative 2 end exponent end fraction space space space space space space equals 5.0 space cross times space 10 to the power of negative 7 end exponent

pK subscript straight a space equals space minus log space space straight K subscript straight a space equals space minus log space left parenthesis 5.0 space cross times space 10 to the power of negative 7 end exponent right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space equals space 7 minus 0.699 space equals space 6.301

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