Question

Calculate the $pH$ at equivalence point of the titration between $0.1MC{H}_{3}COOH\left(25ml\right)$ with $0.05MNaOH{K}_a$ for $C{H}_{3}COOH=1.8\times {10}^{-5}$.

Answer 1

$0.1M$ $C{H}_{3}COOH$ $\left(25ml\right)+0.05M$ $\left(NaOH\right)$

$K_a\left(C{H}_{3}COOH\right)=1.8\times {10}^{-5}$

$p{K}_a=5-log1.8=4.745$

$C{H}_{3}COO{H}^{-}+O{H}^{-}\leftrightharpoons C{H}_{3}CO{O}^{-}+H_{2}O$

$n_{acetic}=\left(0.1\right)\left(0.025\right)=0.00250$

$n_{hydroxide}=\frac{0.10}{2}\frac{20}{1000}=0.00125$

$n_{acetic}$ remains$=0.00125$

also, $0.00125$ Acetic ions.

$pH=p{K}_a+log\frac{conjugatebase}{weakacid}$

$V_{total}=25$ $ml$

$\left[C{H}_{3}CO{O}^{-}\right]=\left[C{H}_{3}COOH\right]=\frac{0.00125}{\frac{25}{1000}}$

$pH=4.745$

$pH=p{K}_a+log\frac{C{H}_{3}CO{O}^{-}}{C{H}_{3}COOH}$

$⟹pH=p{K}_a+0$

$H_{2}A⟷H{A}^{-}+H^{+}$

$k_1=\frac{\left[{HA}^{-}\right]\left[H^{+}\right]}{\left[H_{2}A\right]}$

$H{A}^{-}⟷A^{2-}+H^{+}$

$K_1=\frac{\left[A^{2-}\right]\left[H^{+}\right]}{\left[{HA}^{-}\right]}$

$H_{2}A\rightleftharpoons A^{2-}+2H^{+}$

$K=\frac{\left[A^{2-}\right]{\left[H^{+}\right]}^2}{\left[H_{2}A\right]}=K_1\times K_2$

$K=\left({10}^{-5}\right)(5\times {10}^{-10})$

Overall dissociation Constant,

$K=5\times {10}^{-15}$