JEE Chemistry 2023 Question with Solution

Given: Initial volume is $1 \, \text{L}$ and the HCl solution changes from $\text{pH} = 1$ to $\text{pH} = 2$.

Find: The volume of water to be added in $\text{mL}$.

From the solution:

$$M_1 = 10 \quad (\text{pH} = 1, \text{ concentration of HCl})$$ $$M_2 = 10^{-2} \quad (\text{pH} = 2, \text{ concentration of HCl})$$ $$V_1 = 1 \, \text{L}$$

Using the dilution formula:

$$(M_1 \times V_1) = (M_2 \times V_2)$$ $$10 \times 1 = 10^{-2} \times V_2$$ $$V_2 = 10 \, \text{L} = 10000 \, \text{mL}$$

Thus, water added is

$$10000 - 1000 = 9000 \, \text{mL}$$

Therefore, the required volume of water is $9000$.

Given: The pH changes from $1$ to $2$ for $1 \, \text{L}$ of HCl.

Find: Water added.

Quick Tip: For dilution, the product of concentration and volume remains constant. You can solve it using the dilution equation.

A change in pH from $1$ to $2$ means the acid concentration becomes one-tenth of its initial value. Therefore, the final volume becomes $10$ times the initial volume:

$$V_2 = 10 \, \text{L} = 10000 \, \text{mL}$$

Initial volume is $1000 \, \text{mL}$, so water added is

$$10000 - 1000 = 9000 \, \text{mL}$$

Therefore, the answer is $9000$.