JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 17)
$$\mathrm{K}_{\mathrm{a}}$$ for butyric acid $$\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{COOH}\right)$$ is $$2 \times 10^{-5}$$. The $$\mathrm{pH}$$ of $$0.2 \,\mathrm{M}$$ solution of butyric acid is __________ $$\times 10^{-1}$$. (Nearest integer)
[Given $$\log 2=0.30$$]
Answer
27
Explanation
$\mathrm{K}_{\mathrm{a}}$ of Butyric acid $\Rightarrow 2 \times 10^{-5} \,\mathrm{PKa}=4.7$
$\mathrm{pH}$ of $0.2 \mathrm{M}$ solution,
$$ \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C} $$
$$ \begin{aligned} &=\frac{1}{2}(4 \cdot 7) - \frac{1}{2} \log (0.2) \\\\ &=2.35+0.35=2.7 \end{aligned} $$
$$ \mathrm{pH}=27 \times 10^{-1} $$
$\mathrm{pH}$ of $0.2 \mathrm{M}$ solution,
$$ \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C} $$
$$ \begin{aligned} &=\frac{1}{2}(4 \cdot 7) - \frac{1}{2} \log (0.2) \\\\ &=2.35+0.35=2.7 \end{aligned} $$
$$ \mathrm{pH}=27 \times 10^{-1} $$
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