The $pH$ of $0.004 \,M$ hydrazine solution is $9.7 .$ Calculate its ionization constant $K_{ b }$ and $pK _{ b }$
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,
$K_{ b }=\left[ NH _{2} NH _{3}^{+}\right]\left[ OH ^{\top}\right] /\left[ NH _{2} NH _{2}\right]$
$=\left(5.98 \times 10^{-5}\right)^{2} / 0.004=8.96 \times 10^{-7}$
$p K_{ b }=-\log K_{ b }=-\log \left(8.96 \times 10^{-7}\right)=6.04$
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When $100 \ mL$ of $1.0 \ M \ HCl$ was mixed with $100 \ mL$ of $1.0 \ M \ NaOH$ in an insulated beaker at constant pressure, a temperature increase of $5.7^{\circ} C$ was measured for the beaker and its contents (Expt. $1$). Because the enthalpy of neutralization of a strong acid with a strong base is a constant $\left(-57.0 \ kJ \ mol ^{-1}\right)$, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. $2$), $100 \ mL$ of $2.0 \ M$ acetic acid $\left(K_a=2.0 \times 10^{-5}\right)$ was mixed with $100 \ mL$ of $1.0 M \ NaOH$ (under identical conditions to Expt. $1$) where a temperature rise of $5.6^{\circ} C$ was measured.
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$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$
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$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$
Give the answer question $1$ and $2.$
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$1.$ Enthalpy of dissociation (in $kJ mol ^{-1}$ ) of acetic acid obtained from the Expt. $2$ is
$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$
$2.$ The $pH$ of the solution after Expt. $2$ is
$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$
Give the answer question $1$ and $2.$
- [IIT 2015]
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