What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0

Question

What is the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.507 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107. Express the pH numerically to three decimal places. pH = 6.345

Part B What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places.

Part C What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places.

Explanation / Answer

A) pH = pKa + log [NaA] / [HA]

pKa = - log Ka = - log [5.66 x10-7] = 6.25

[HA] = 0.405/2 = 0.2025 M

[NaA] = 0.507 / 2 = 0.2535 M

pH = 6.25 + log [0.2535] / [0.2025]

pH = 6.347

part B:

0.15 / 2 = 0.075 M HCl added

[NaA] = 0.2535 - 0.075 = 0.1785 M

[HA] = 0.2025 + 0.075 = 0.2775 M

pH = 6.25 + log [0.1785] /[0.2775]

pH = 6.058

part C:

0.195 /2 = 0.0975 M NaOH added

[NaA] = 0.2535 + 0.0975 = 0.351 M

[HA] = 0.2025 - 0.0975 = 0.105 M

pH = 6.25 + log [0.351] / [0.105]

pH = 6.774

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