JEE MAIN - Chemistry (2023 - 25th January Morning Shift - No. 15)
A litre of buffer solution contains 0.1 mole of each of NH$$_3$$ and NH$$_4$$Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____________ $$\times$$ 10$$^{-3}$$ (Nearest integer)
[Given : $$\mathrm{pK_b(NH_3)=4.745}$$
$$\mathrm{\log2=0.301}$$
$$\mathrm{\log3=0.477}$$
$$\mathrm{T=298~K]}$$
Answer
9079
Explanation
In resultant solution
$$ \begin{aligned} & \mathrm{n}_{\mathrm{NH}_3}= 0.1-0.02=0.08 \\\\ & \mathrm{n}_{\mathrm{NH}_4 \mathrm{Cl}}= \mathrm{n}_{\mathrm{NH}_4^{+}}=0.1+0.02=0.12 \\\\ & \mathrm{pOH}= \mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\\\ &=4.745+\log \frac{0.12}{0.08} \\\\ &=4.745+\log \frac{3}{2} \\\\ &=4.745+0.477-0.301 \\\\ & \mathrm{pOH}=4.921 \\\\ & \mathrm{pH}=14-\mathrm{pH} \\\\ &=9.079 = 9079\times 10^{-3} \end{aligned} $$
$$ \begin{aligned} & \mathrm{n}_{\mathrm{NH}_3}= 0.1-0.02=0.08 \\\\ & \mathrm{n}_{\mathrm{NH}_4 \mathrm{Cl}}= \mathrm{n}_{\mathrm{NH}_4^{+}}=0.1+0.02=0.12 \\\\ & \mathrm{pOH}= \mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\\\ &=4.745+\log \frac{0.12}{0.08} \\\\ &=4.745+\log \frac{3}{2} \\\\ &=4.745+0.477-0.301 \\\\ & \mathrm{pOH}=4.921 \\\\ & \mathrm{pH}=14-\mathrm{pH} \\\\ &=9.079 = 9079\times 10^{-3} \end{aligned} $$
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