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Using a 0.25 M phospate buffer with a pH of 7.2 you add 0.74 mL go...

Using a 0.25 M phospate buffer with a pH of 7.2 you add 0.74 mL go 0.50 M HCl to 54 mL of the buffer. What is the new pH of the solution? Enter answer in 3 significant figures

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Answer #1

Since pH = pKa, the concentration of the buffer components are the same:

n Salt = n Acid = M * V / 2 = 0.25 M * 0.054 L / 2 = 0.00675 mol

The added moles of HCl are calculated:

n HCl = 0.5 M * 0.00074 L = 0.00037 mol

The HCl reacts with the salt and forms acid, the new pH is calculated:

pH = pKa + log (n Salt / n Acid) = 7.20 + log (0.00675 - 0.00037 / 0.00675 + 0.00037) = 7.15

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