We’re being asked to calculate the pH of buffer composed of H_{2}PO_{4}^{-}_{(aq)} and HPO_{4}^{2-}_{(aq)}. To calculate for pH, we’re going to use the **Henderson-Hasselbalch equation**:

$\overline{){\mathbf{p}}{\mathbf{H}}{\mathbf{=}}{\mathbf{p}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{l}}{\mathbf{o}}{\mathbf{g}}\frac{\left[\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{j}\mathbf{u}\mathbf{g}\mathbf{a}\mathbf{t}\mathbf{e}\mathbf{}\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}\right]}{\left[\mathbf{w}\mathbf{e}\mathbf{a}\mathbf{k}\mathbf{}\mathbf{a}\mathbf{c}\mathbf{i}\mathbf{d}\right]}}$

We know that

$\overline{){\mathbf{p}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{l}}{\mathbf{o}}{\mathbf{g}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{a}}}}$

So first have to figure out which K_{a} value we’re going to use for our calculation. Phosphoric acid (H_{3}PO_{4}) is a weak triprotic acid, meaning it can donate three protons (H^{+}) and it will have three equilibrium reactions.

**Removing one H ^{+} per step in H_{3}PO_{4}:**

Phosphoric acid is a triprotic acid (Ka_{1} = 6.9 x 10^{-3}, Ka_{2} = 6.2x 10^{-8}, and Ka_{3} = 4.8 x10^{-13}). To find the pH of a buffer composed of H_{2}PO_{4}^{-} (aq) and HPO_{4}^{2-} (aq), which pK_{a} value would you use in the Henderson-Hasselbalch equation?

Calculate the pH of a buffer solution obtained by dissolving 13.0 g of KH_{2}PO_{4}(s) and 26.0 g of Na_{2}HPO_{4}(s) in water and then diluting to 1.00 L.

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Our tutors have indicated that to solve this problem you will need to apply the Polyprotic Acid concept. You can view video lessons to learn Polyprotic Acid. Or if you need more Polyprotic Acid practice, you can also practice Polyprotic Acid practice problems.

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Our tutors rated the difficulty of*Phosphoric acid is a triprotic acid (Ka1 = 6.9 x 10-3, Ka2 =...*as high difficulty.

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