Question

A quantity of NaOH of mass 1.1g is dissolved in water and made up to 250cm³ in a graduated flask. A 20cm³ of the solution neutralizes 20.4cm³ of a solution of acetic acid of ph=2.85

Calculate

a. Number of moles of NaOH in the 250cm³ solution

b. Number of moles of acetic acid in 1dm³ of the solution

Answer 1

a. Number of moles of NaOH in the 250cm³ solution

$$\nu=m/M$$

$\nu(NaOH)=m(NaOH)/M(NaOH)=1.1 g/(40g/mol)=0.0275mol$

b. Number of moles of acetic acid in 1dm³ of the solution

$CH_3COOH=>H^++CH_3COO^-$

$[H^+]=[CH_3COO^-]$

pH = - lg [H⁺] => [H⁺]=10^-pH = 10^-2.85 = 0.001413 mol/L

$K_a(CH_3COOH) =1.75\cdot 10^-5=[H^+][CH_3COO^-]/[CH_3COOH]=>$

$[CH_3COOH]=[H^+][CH_3COO^-]/K_a=(0.001413 )^2/(1.75\cdot 10^-5)=0.114 mol/L$

CH₃COOH is a week acid, therefore:

$[CH_3COOH]\approx C(CH_3COOH)=0.114 mol/L$

$$C=\nu/V=>\nu=C\cdot V=0.114mol/L\cdot 1L=0.114 mol$$