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$100 \mathrm{~mL}$ of $2 \mathrm{M}$ of formic acid $\left(\mathrm{p} K_a=3.74\right)$ is neutralise by $\mathrm{NaOH}$, at the equivalence point $\mathrm{pH}$ is
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The correct answer is:
$8.87$
$8.87$
Sodium formate is present at the equivalence point. It is the salt of weak acid $+$ strong base. So, final solution will be basic in nature.
As we know
$\mathrm{pH}=7+\frac{\mathrm{p} K_a}{2}+\frac{\log C}{2}$
where, $C$ is concentration of salt.
Total volume of solution $=100+100=200 \mathrm{~mL}$
Concentration of salt $(C)=2 \times \frac{100}{200}=1 \mathrm{M}$
$\mathrm{pH}=7+\frac{3.74}{2}+\frac{\log [1]}{2}$
$\mathrm{pH}=7+1.87+0$
$\mathrm{pH}=8.87$
As we know
$\mathrm{pH}=7+\frac{\mathrm{p} K_a}{2}+\frac{\log C}{2}$
where, $C$ is concentration of salt.
Total volume of solution $=100+100=200 \mathrm{~mL}$
Concentration of salt $(C)=2 \times \frac{100}{200}=1 \mathrm{M}$
$\mathrm{pH}=7+\frac{3.74}{2}+\frac{\log [1]}{2}$
$\mathrm{pH}=7+1.87+0$
$\mathrm{pH}=8.87$
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