JEE Chemistry 2023 Question with Solution

Given: A buffer contains $0.1$ mole each of $\mathrm{NH_3}$ and $\mathrm{NH_4Cl}$ in $1$ litre. $0.02$ mole of $\mathrm{HCl}$ is added.

Find: The value of pH expressed as _____ $\times 10^{-3}$.

In resultant solution,

$$n_{\mathrm{NH_3}} = 0.1 - 0.02 = 0.08$$ $$n_{\mathrm{NH_4^+}} = 0.1 + 0.02 = 0.12$$

Using the buffer relation for a weak base,

$$\mathrm{pOH} = \mathrm{p}K_b + \log \frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]}$$ $$= 4.745 + \log \frac{0.08}{0.12}$$ $$= 4.745 + \log \frac{2}{3}$$

Using the given logarithms,

$$\log \frac{2}{3} = \log 2 - \log 3 = 0.301 - 0.477 = -0.176$$

Therefore,

$$\mathrm{pOH} = 4.745 - 0.176 = 4.921$$

Now,

$$\mathrm{pH} = 14 - 4.921 = 9.079$$

So the pH is $9.079$, hence the required nearest integer in the form $\text{pH} = 9.079 \times 10^{-3}$ is $9079$ as concluded in the provided solution.

Therefore, the final answer is $9079$.

Given: Basic buffer of $\mathrm{NH_3/NH_4Cl}$ with initial moles $0.1$ and $0.1$. Added acid = $0.02$ mole $\mathrm{HCl}$.

Find: Numerical value asked in the question.

The added strong acid consumes the weak base and converts it into its conjugate acid:

$$\mathrm{NH_3} + \mathrm{HCl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl^-}$$

Hence base decreases by $0.02$ and conjugate acid increases by $0.02$.

For a basic buffer,

$$\mathrm{pH} = 14 - \mathrm{p}K_b + \log \frac{[\text{Base}]}{[\text{Acid}]}$$

Substituting,

$$\mathrm{pH} = 14 - 4.745 + \log \frac{0.08}{0.12}$$ $$= 14 - 4.745 + \log \frac{2}{3}$$ $$= 14 - 4.745 + 0.301 - 0.477$$ $$= 14 - 4.921 = 9.079$$

the solution reports the answer in the requested transformed form as $9079$.

Therefore, the correct numerical answer is $9079$.