Answer
Buffer A: 3.7
Buffer B: 3.8
Work Step by Step
The acid equilibrium constant for formic acid is stated in the Appendix section of the book: $Ka=1.8\times10^{-4}$.
I. Buffer A:
$CH_{3}COONa -> Na^{+} + CH_{3}COO^{-}$
Because sodium formate dissociates completely in water, initially $[Na^{+}]=[CH_{3}COO^{-}]=1M$
Then, we construct a table describing the initial and equilibrium concentrations of molecules based on formic acid's dissociation.
Reaction: $HCOOH + H_{2}O H_{3}O^{+} + COOH^{-}$
Initial:............$1$ M.................................................$0$ M.............$1$ M
Change:....$-x$ M...............................................$x$ M.............$x$ M
Equilibria: $1-x$ M.........................................$x$ M.............$1+x$ M
Therefore:
$Ka=\frac{[H_{3}O^{+}][COOH^{-}]}{[HCOOH]}=\frac{x(1+x)}{1-x}=1.8\times10^-4$
(use calculator) $x=[H^{+}]\approx1.8\times10^{-4} $ M
$pH=-log[H^{+}]=-log(1.8\times10^{-4})=3.7$
II. Buffer B:
Similar to the above, initially $[Na^{+}]=[CH_{3}COO^{-}]=0.01M$
Reaction: $HCOOH + H_{2}O H_{3}O^{+} + COOH^{-}$
Initial:..........$0.01$ M...........................................$0$ M.............$0.01$ M
Change:....$-x$ M...............................................$x$ M.............$x$ M
Equilibria: $0.01-x$ M..................................$x$ M.............$0.01+x$ M
Therefore:
$Ka=\frac{[H_{3}O^{+}][COOH^{-}]}{[HCOOH]}=\frac{x(0.01+x)}{0.01-x}=1.8\times10^-4$
(use calculator) $x=[H^{+}]\approx1.7\times10^{-4} $ M
$pH=-log[H^{+}]=-log(1.7\times10^{-4})=3.8$
