Question

If the equilibrium constant of the reaction of weak acid $HA$ with a strong base is ${10}^9$, then the $pH$ of $0.1M$ $NaA$ solution will be:

Answer 1

$HA+NaOH\rightleftharpoons NaA+H_{2}O$

$K=10$

For $NaA+H_{2}O\rightleftharpoons NaOH+HA;K={10}^9$

$\frac{(}{NaOH}HA\left)\right(\left[NaH\right])={10}^{-}9$

$\frac{x^2}{0.1-x}={10}^{-}9$

$x^2={10}^{-}9\ast 0.1$

$x={10}^{-}5$

$\left[OH\right]=\left[NaOH\right]={10}^{-}5M$

$p^H=14-POH=14-(-log{10}^{-}5)=14-5=9$