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

pH = 2.462.

Step-by-step 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]=(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=([NO_2^-][H_3O^+])/([HNO_2])\\\\7.1x10^(-4)=(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!