The $pH$ of $0.004 \,M$ hydrazine solution is $9.7 .$ Calculate its ionization constant $K_{ b }$ and $pK _{ b }$
$NH _{2} NH _{2}+ H _{2} O \rightleftharpoons NH _{2} NH _{3}^{+}+ OH ^{-}$
From the $pH$ we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:
$\left.| H ^{+}\right]=$ antilog $(- pH )$
$=$ antilog $(-9.7)=1.67 \times 10^{-10}$
$\left[ OH ^{-}\right]=K_{ w } /\left[ H ^{+}\right] =1 \times 10^{-14} / 1.67 \times 10^{-10} $
$=5.98 \times 10^{-5}$
The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to $0.004 \,M$ Thus,
$K_{ b }=\left[ NH _{2} NH _{3}^{+}\right]\left[ OH ^{\top}\right] /\left[ NH _{2} NH _{2}\right]$
$=\left(5.98 \times 10^{-5}\right)^{2} / 0.004=8.96 \times 10^{-7}$
$p K_{ b }=-\log K_{ b }=-\log \left(8.96 \times 10^{-7}\right)=6.04$
Dissociation constat of weak acid $HA$ is $1.8 \times {10^{ - 4}}$ calculate Dissociation constant of its conjugate base ${A^ - }$
Values of dissociation constant, $K_a$ are given as follows
Acid | $K_a$ |
$HCN$ | $6.2\times 10^{-10}$ |
$HF$ | $7.2\times 10^{-4}$ |
$HNO_2$ | $4.0\times 10^{-4}$ |
Correct order of increasing base strength of the base $CN^-,F^-$ and $NO_2^-$ will be
$K_b$ for $NH_4OH$ is $1.8\times 10^{-5}.$ The $[\mathop O\limits^\Theta H]$ of $0.1\,M\,NH_4OH$ is
Calculate $pH$ of $0.02$ $mL$ $ClC{H_2}COOH$. Its ${K_a} = 1.36 \times {10^{ - 3}}$ calculate its $pK_{b}$,
Calculate the $pH$ of the solution in which $0.2 \,M\, NH _{4} Cl$ and $0.1 \,M\, NH _{3}$ are present. The $pK _{ b }$ of ammonia solution is $4.75$