## Chemistry and Chemical Reactivity (9th Edition)

$pH=11.29$
At the equivalence point, initially all the acid is consumed, so the volume of NaOH required is: $25.0\ mL\times 0.090\ M=0.108\ M\times V\rightarrow V=20.83\ mL$ $[A^-]=0.090\ M\times 25.0\ mL/(25.0\ mL+20.83\ mL)=0.049\ M$ Then, calculate the pH by the equilibrium: $A^-+H_2O\leftrightarrow HA+OH^-$ $Kw/Ka=[HA][OH^-]/[A^-]$ $10^{-14}/(1.3\times10^{-10})=[OH^-]^2/0.049\ M$ $[OH^-]=1.941\times10^{-3}\ M$ $pH=pKw-pOH$ $pH=14+\log(1.941\times10^{-3})$ $pH=11.29$