Answer
$pH=8.21$
Work Step by Step
$CO_2H_2+OH^- \rightarrow CO_2H^-+H_2O$
First, assume that all the hydroxide is consumed:
$n_{OH^-}=50.0\ mL\times 0.070\ M=3.5\ mmol$
$n_{CO_2H_2}=25.0\ mL\times 0.14\ M=3.5\ mmol$
Since the number of moles is equal, all the initial formic acid is present as formate. Then, calculate the pH by the equilibrium:
$CO_2H^-+H_2O\leftrightarrow CO_2H_2+OH^-$
$K_w/K_a=[CO_2H_2][OH^-]/[CO_2H^-]$
$[CO_2H^-]_0=3.5\ mmol / (50+25)\ mL=0.0467\ M$
$10^{-14}/1.77\times10^{-4}=x\times x/(0.0467-x)$
$x=[OH^-]=1.624\times10^{-6}$
$pH=pKw-pOH$
$pH=14+\log(1.624\times10^{-6})$
$pH=8.21$
