Question

A solution of hydrochloric acid has a pH of 1.8. A 1.0 g sample of sodium hydroxide is dissolved into 272.6 mL of this hydrochloric acid solution. Calculate the POH of the resulting solution, assuming no volume change

Answer 1

$HCl(aq) + NaOh (aq) = NaCl(aq) + H_2O (l)$

1) moles of NaOH:

$1.0 g NaOH (\frac{1 mole NaOH}{40.0 g NaOH})= 0.025 mol NaOH$

2) moles of HCl

$pH = -log_{10}[H^+]$

$[H^+]= 10^{-pH} = 10^{-1.8} = 0.01585 mol/L$

$272.6 mL (\frac{1 L}{1000mL})(\frac{0.01585 mol HCl}{1 L})=0.004321 mol$

3) As mole ratio n(HCl):n(NaOH) = 1:1, then moles of both NaOh and HCl consumed in the reaction are the same. As limiting reactant is HCl (0.004321<0.025), then 0.004321 moles of NaOH were consumed in the reaction.

Moles of NaOH left:

$n(NaOH) = 0.025-0.004321 = 0.02068 mol$

concentration of NaOH solution is:

$c(NaOH) = \frac{moles NaOh}{volume of solution} = \frac{0.02068 mol}{0.2726L}=0.07586 mol/L$

4) $pOH = -log_{10}[OH^-]= -log_{10}(0.07586)= 1.1$