Answer:
pH = 2.462.
Explanation:
Hello there!
In this case, according to the reaction between nitrous acid and potassium hydroxide:
[tex]HNO_2+KOH\rightarrow KNO_2+H_2O[/tex]
It is possible to compute the moles of each reactant given their concentrations and volumes:
[tex]n_{HNO_2}=0.02000L*0.1000mol/L=2.000x10^{-3}mol\\\\n_{KOH}=0.1000mol/L*0.01327L=1.327x10^{-3}mol[/tex]
Thus, the resulting moles of nitrous acid after the reaction are:
[tex]n_{HNO_2}=2.000x10^{-3}mol-1.327x10^{-3}mol=6.73x10^{-4}mol[/tex]
So the resulting concentration considering the final volume (20.00mL+13.27mL) is:
[tex][HNO_2]=\frac{6.73x10^{-4}mol}{0.01327L+0.02000L} =0.02023M[/tex]
In such a way, we can write the ionization of this weak acid to obtain:
[tex]HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+[/tex]
So we can set up its equilibrium expression to obtain x as the concentration of H3O+:
[tex]Ka=\frac{[NO_2^-][H_3O^+]}{[HNO_2]}\\\\7.1x10^{-4}=\frac{x^2}{0.02023M-x}[/tex]
Next, by solving for the two roots of x, we get:
[tex]x_1=-0.004161M\\\\x_2=0.003451M[/tex]
Whereas the correct value is 0.003451 M. Finally, we compute the resulting pH:
[tex]pH=-log(0.003451)\\\\pH=2.462[/tex]
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