A chemist prepares 1.0 L of a buffer solution by preparing an aqueous solution containing 1.00 mol of H2PO4−, then adding 0.27 moles of KOH to convert the acid into its conjugate base. What is the final pH of the solution? The Ka for H2PO4− is 6.2 x 10−8; the pKa is 7.20.

pH of solution is 6.77.

Explanation:

H₂PO₄⁻ reacts with KOH thus:

H₂PO₄⁻ + KOH → HPO₄²⁻ + H₂O + K⁺

If 1.00 mol of H₂PO₄⁻ reacts with 0.27mol of KOH, moles of HPO₄²⁻ produced are 0.27mol and moles of H₂PO₄⁻ that remains are 1.00mol - 0.27mol = 0.73mol.

It is possible to find pH of this buffer using H-H equation:

pH = pka + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]

Where pKa is 7.20

[HPO₄²⁻] = 0.27mol / 1.0L = 0.27M

[H₂PO₄⁻] = 0.73mol / 1.0L = 0.73M

Replacing:

pH = 7.20 + log₁₀ [0.27M] / [0.73M]

pH = 6.77

pH of solution is 6.77

the formula would be mgso4

mgso[subscript]4

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