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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.
(Consider heat capacity of all solutions as $4.2 J g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g mL ^{-1}$ )
$1.$ 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 $2.$
$(A,B)$
$(B,D)$
$(B,C)$
$(A,C)$
Solution
$1.$ $HCl + NaOH \longrightarrow NaCl + H _2 O$
$n =100 \times 1=100 m \text { mole }=0.1 mole$
Energy evolved due to neutralization of $HCl$ and $NaOH =0.1 \times 57=5.7 kJ =5700$ Joule
Energy used to increase temperature of solution $=200 \times 4.2 \times 5.7=4788$ Joule
Energy used to increase temperature of calorimeter $=5700-4788=912 Joule$
$ms . \Delta t =912$
$m . s \times 5.7=912$
$ms =160 \text { Joule } /{ }^{\circ} C \text { [Calorimeter constant] }$
Energy evolved by neutralization of $CH _5 COOH$ and $NaOH$
$=200 \times 4.2 \times 5.6+160 \times 5.6=5600 \text { Joule }$
So energy used in dissociation of $0.1 mole CH _3 COOH =5700-5600=100$ Joule Enthalpy of dissociation $=1 kJ / mole$
$2.$ $CH _3 COOH =\frac{1 \times 100}{200}=\frac{1}{2}$
$CH _3 CONa =\frac{1 \times 100}{200}=\frac{1}{2}$
$pH = pK _{ a }+\log \frac{[\text { salt }]}{[\text { acid }]}$
$pH =5-\log 2+\log \frac{1 / 2}{1 / 2}$
$pH =4.7$