## Chemistry and Chemical Reactivity (9th Edition)

$pH=8.21$
$CO_2H_2+OH^- \rightarrow CO_2H^-+H_2O$ 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: $CO_2H^-+H_2O\leftrightarrow CO_2H_2+OH^-$ $K_w/K_a=[CO_2H_2][OH^-]/[CO_2H^-]$ $[CO_2H^-]_0=3.5\ mmol / (50+25)\ mL=0.0467\ M$ $10^{-14}/1.77\times10^{-4}=x\times x/(0.0467-x)$ $x=[OH^-]=1.624\times10^{-6}$ $pH=pKw-pOH$ $pH=14+\log(1.624\times10^{-6})$ $pH=8.21$