Question

A 555 mL of 0.500 M lactate buffer (Ka = 1.38 x 104), pH 4.0 is to be prepared. How many moles of NaOH is needed to prepare the buffer?

Answer 1

K_a = 1.38×10^-4

pK_a =10^0.139 × 10^-4 = 4-0.139 = 3.861

pH = pK_a + log$\frac{salt}{acid}$

4 = 3.861 + log$\frac{salt}{0.555×0.5}$

1.38 = $\frac{salt}{0.2775}$

moles of salt = 1.38×0.2775 = 0.38295M