Question

How many grams of sodium benzoate (NaC7H5O2) is needed to make a 175.0mL buffer solution containing 0.15 M benzoic acid and a pH of 3.91?

Answer 1

Solution.

$pH = pKa - lg(\frac{C(acid)}{C(base)})$

$Acid \ is \ C7H6O2$

$Base \ is \ C7H5O2^-$

$C(base) = C(acid) \times 10^{pH-pKa}$

C(base) = 0.075 M

$m(C7H5O2Na) = C(base) \times M(C7H5O2Na) \times V(solution)$

m(C7H5O2Na) = 1.90 g

Answer:

m(C7H5O2Na) = 1.90 g