Question

Dissociation constants of $C{H}_{3}COOH$ and $N{H}_{4}OH$ in aqueous solution are ${10}^{-5}$. If pH of a $C{H}_{3}COOH$ solution is $3$. What will be the pH of $N{H}_{4}OH$?

• A. $3.0$
• B. $4.0$
• C. $10.0$
• D. $11.0$

Answer 1

Answer: (D)

$11.0$

$C{H}_{3}COOH+H_{2}O\to H_3{O}^{+}+C{H}_{3}CO{O}^{-}$

$K_a=\frac{C{\alpha }^2}{1-\alpha }$

$C{\alpha }^2=K_4$

$P^H=3.5$

$-\log H+=3$

$(H+)={10}^{-3}$

$c\alpha ={10}^{-3}$

$\alpha =\frac{{10}^{-5}}{{10}^{-3}}={10}^{-2}$

since both have same dissociation

constants so $\left[O{H}^{-}\right]={10}^{-3}$

$P^H=14+\log \left[O{H}^{+}\right]$

$P^H=14+\log {10}^{-3}=11$