Question

At a pH of 7.40, what is the ratio of the molar concentrations of H2_{2}PO4_{4}^{-} to H3_{3}PO4_{4}?

Solution

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Given data:

pH=7.40\text{pH}=7.40

Ka(H3PO4)=7.1×103 mol/dm3K_{a}(\text{H}_{3}\text{PO}_{4})=7.1 \times 10^{-3} \ \text{mol/dm}^{3}

We have to determine the molar ratio of hydrogenphosphate anion to phosphate acid.

Buffer solutions consists of a week acid and its conjugate base. Since the hydrogenphosphate anion is a conjugated base of phosphate acid, we can assume that we have a buffer solution here.

Buffer solutions follow the Handerson's Hasselbach equation:

pH=pKa+log[A][HA](1)\text{pH}=\text{pK}_{a}+ \log\dfrac{[\text{A}^{-}]}{[\text{HA}]} \tag{1}

So, knowing the constant dissociation at the pH value, we can determine the molar ratio of acid and its conjugate base.

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