JEE Main 2023 — Ionic Equilibrium Question with Solution

We can use the formula for pH to calculate the concentration of hydrogen ions in each solution:

$$\mathrm{pH}=-{\log }_{10}[{\mathrm{H}}^{+}]$$

For the first solution, we have:

$$1=-{\log }_{10}[{\mathrm{H}}^{+}]$$

Solving for $[{\mathrm{H}}^{+}]$, we get:

$$[{\mathrm{H}}^{+}]=0.1\mathrm{M}$$

For the second solution, we want:

$$2=-{\log }_{10}[{\mathrm{H}}^{+}]$$

Solving for $[{\mathrm{H}}^{+}]$, we get:

$$[{\mathrm{H}}^{+}]=0.01\mathrm{M}$$

To dilute the first solution to the desired concentration, we can use the dilution equation:

$$C_1{V}_1=C_2{V}_2$$

where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume.

We know $C_1=0.1\mathrm{M}$, $C_2=0.01\mathrm{M}$, and $V_1=1\mathrm{L}$. Solving for $V_2$, we get:

$$V_2=\frac{C_1{V}_1}{C_2}=\frac{(0.1\mathrm{M})(1\mathrm{L})}{0.01\mathrm{M}}=10\mathrm{L}$$

Therefore, we need to add $10-1=9$ liters of water to the initial solution to obtain the desired pH. Converting liters to milliliters, we get:

$$9\mathrm{L}\times \frac{1000\mathrm{mL}}{1\mathrm{L}}=9000\mathrm{mL}$$

So the volume of water needed is 9000 mL or 9,000 mL (nearest integer).