Question

A 5.60 g sample of a weak acid with
Ка 1.3 x 10-4 was combined with 5.00 mL of
5.80 mol L-1 NaOH and the resulting solution
was diluted to 750 mL. The measured pH of the
solution was 4.35.
What is the molar mass of the weak acid

Answer 1

Answer:

pKa = - log Ka = - log (1.3 * 10^-4) = 3.886 5.20 mL of 6.20 M NaOH = 0.00520 L * 6.20 mole…

Explanation: