Question

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

Answer 1

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

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:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

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:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

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:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

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