Answer:
1. pH = 1.23.
2. [tex]H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)[/tex]
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
[tex]H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+[/tex]
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
[tex]pH=pKa+log(\frac{[base]}{[acid]} )[/tex]
Whereas the pKa is:
[tex]pKa=-log(Ka)=-log(5.90x10^{-2})=1.23[/tex]
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
[tex]pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23[/tex]
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
[tex]n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol[/tex]
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
[tex]H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)[/tex]
Which is also shown in net ionic notation.
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