A buffer is prepared by adding 0.450 moles of H2PO4^- and 0.450 moles of HPO4 ^2

Question

A buffer is prepared by adding 0.450 moles of H2PO4^- and 0.450 moles of HPO4 ^2- to 1.000 L of water. The pH of the solution is 7.210 Ka(H2PO4 ^-) = 6.2 x 10^ -8. Calculate the pH of this buffer solution after adding 38.95 mL of 5.75 M HCl. (assume that the final volume is still 1.000 L) A buffer is prepared by adding 0.450 moles of H2PO4^- and 0.450 moles of HPO4 ^2- to 1.000 L of water. The pH of the solution is 7.210 Ka(H2PO4 ^-) = 6.2 x 10^ -8. Calculate the pH of this buffer solution after adding 38.95 mL of 5.75 M HCl. (assume that the final volume is still 1.000 L) Ka(H2PO4 ^-) = 6.2 x 10^ -8. Calculate the pH of this buffer solution after adding 38.95 mL of 5.75 M HCl. (assume that the final volume is still 1.000 L)

Explanation / Answer

The concentration of the weak acid H2PO4-(aq) is, [H2PO4-(aq)] = 0.450 mol / 1L = 0.450 M

The concentration of the salt, HPO42-(aq) is, [HPO42-(aq)] = 0.450 mol /1L = 0.450 M

Moles of HCl added to the buffer solution = MxV(L) = 5.75 mol/L x 0.03895 L = 0.224 mol HCl

Now 0.224 mol HCl will react with 0.224 mol HPO42-(aq) to form 0.224 mol H2PO4-(aq).

Hence moles of HPO42-(aq) remained in the final solution = 0.450 mol - 0.224 mol = 0.226 mol

moles of H2PO4-(aq) remained in the final solution = 0.450 mol + 0.224 mol = 0.674 mol

Ka = 6.2x10-8 => pKa = - logKa = 7.210

Applying Hendersen equation:

pH = pKa + log ([HPO42-(aq)] / [H2PO4-(aq)])

=> pH = pKa + log (moles of HPO42-(aq) / moles of H2PO4-(aq)) [ Since V remains same for both)

=> pH = 7.210 + log(0.226 mol / 0.674 mol)

=> pH = 6.735 (answer)

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