Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M HNO2(aq) with 0.1000 M KOH(aq) after 13.27 mL of the base have been added. Ka of nitrous acid = 7.1 x 10-4.

Answer 1

Answer:

pH = 2.462.

Explanation:

Hello there!

In this case, according to the reaction between nitrous acid and potassium hydroxide:

$HNO_2+KOH\rightarrow KNO_2+H_2O$

It is possible to compute the moles of each reactant given their concentrations and volumes:

$n_{HNO_2}=0.02000L*0.1000mol/L=2.000x10^{-3}mol\\\\n_{KOH}=0.1000mol/L*0.01327L=1.327x10^{-3}mol$

Thus, the resulting moles of nitrous acid after the reaction are:

$n_{HNO_2}=2.000x10^{-3}mol-1.327x10^{-3}mol=6.73x10^{-4}mol$

So the resulting concentration considering the final volume (20.00mL+13.27mL) is:

$[HNO_2]=\frac{6.73x10^{-4}mol}{0.01327L+0.02000L} =0.02023M$

In such a way, we can write the ionization of this weak acid to obtain:

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

So we can set up its equilibrium expression to obtain x as the concentration of H3O+:

$Ka=\frac{[NO_2^-][H_3O^+]}{[HNO_2]}\\\\7.1x10^{-4}=\frac{x^2}{0.02023M-x}$

Next, by solving for the two roots of x, we get:

$x_1=-0.004161M\\\\x_2=0.003451M$

Whereas the correct value is 0.003451 M. Finally, we compute the resulting pH:

$pH=-log(0.003451)\\\\pH=2.462$

Best regards!