Answer
$1.29*10^{-5}$
Work Step by Step
Because $11.05ml/22.10ml=1/2$, we can conclude that there is half the quantity of NaOH required to reach the end point.
$NaOH + HPr -> NaPr + H_{2}O
n_{NaOH}=n_{NaPr}=\frac{1}{2}*1.11*10^{-3}=5.56*10^{-4}(mol)$
Because it is exactly half comparing to the end point, the number of mole HPr remaining is also equal to the mole of HPr reacted, which is $5.56*10^{-4}mol$.
$C_{HPr}=C_{NaPr}=5.56*10^{-4}mol/0.01105l=0.0503M$
$NaPr -> Na^{+}+Pr^{-}$
$0.0503M ->->0.0503M$
With the pH value, we can calculate [H+], thus be able to calculate $K_{a}$.
pH=4.89
$-log[H+]=4.89$
$[H+]=10^{-4.89}=1.29*10^{-5}M$
$HPr + H_{2}O -> H_{3}O^{+}+Pr^{-}$
$[HPr]=0.0503M-1.29*10^{-5}M\approx0.0503M$
$[Pr-]=1.29*10^{-5}M+0.0503M\approx0.0503M$
$K_{a}=\frac{[H+][Pr-]}{[HPr]}=\frac{1.29*10^{-5}*0.0503}{0.0503}=1.29*10^{-5}$
