Answer
$pH=11.29$
Work Step by Step
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:
$A^-+H_2O\leftrightarrow HA+OH^-$
$Kw/Ka=[HA][OH^-]/[A^-]$
$10^{-14}/(1.3\times10^{-10})=[OH^-]^2/0.049\ M$
$[OH^-]=1.941\times10^{-3}\ M$
$pH=pKw-pOH$
$pH=14+\log(1.941\times10^{-3})$
$pH=11.29$
