Question

A buffer solution is prepared by dissolving 1.50gram of benzoic acid HC7H5O2 and sodium benzoate NaC7H5O2 in 150 mL of Solution a. what is the PH of this buffer Solution? ( ka =6.3×10^-5)

Answer 1

mass of Benzoic acid = 1.50g

number of moles of Benzoic acid = 1.50/122.12 = 0.0123 moles

molar concentration [HA] = 0.0123/150 × 1000 = 0.082 M

Ka = 6.3 × 10^-5

pKa = 4.2

from,

$pK_a = pH - \log( \dfrac{A^-}{HA})$

$pH = pK_a + \log( \dfrac{A^-}{HA})$

$pH =4.2 + \log(\dfrac{A^-}{0.082})$