JEE Chemistry 2025 Question with Solution

Given: pH = $10.0$, volume = $1.0 \, \text{L}$, molar mass of Mg(OH)$_2$ = $58 \, \text{g mol}^{-1}$.

Find: the mass of Mg(OH)$_2$ required in mg.

First, calculate pOH:

$$pOH = 14.0 - 10.0 = 4.0$$

Then the hydroxide ion concentration is:

$$[\text{OH}^-] = 10^{-pOH} = 10^{-4.0} = 1.0 \times 10^{-4} \, \text{M}$$

Mg(OH)$_2$ dissociates completely as:

$$\text{Mg(OH)}_2 \rightarrow \text{Mg}^{2+} + 2\text{OH}^-$$

So, if the concentration of Mg(OH)$_2$ is $[\text{Mg(OH)}_2]$, then:

$$2[\text{Mg(OH)}_2] = [\text{OH}^-]$$

Hence,

$$[\text{Mg(OH)}_2] = \frac{1.0 \times 10^{-4}}{2} = 0.5 \times 10^{-4} \, \text{M}$$

For $1.0 \, \text{L}$ solution, moles required are:

$$\text{moles of Mg(OH)}_2 = 0.5 \times 10^{-4} \times 1.0 = 0.5 \times 10^{-4} \, \text{mol}$$

Now convert moles to mass:

$$\text{mass} = 0.5 \times 10^{-4} \times 58 = 2.90 \times 10^{-3} \, \text{g}$$

Converting to mg:

$$2.90 \times 10^{-3} \, \text{g} = 2.90 \, \text{mg}$$

Rounded to the nearest integer, the value of $x$ is $3$.

Given: pH = $10.0$.

Find: mass of Mg(OH)$_2$ in mg.

Use pH to get pOH:

$$pOH = 4$$

So,

$$[\text{OH}^-] = 10^{-4} \, \text{mol L}^{-1}$$

Since one mole of Mg(OH)$_2$ gives two moles of $\text{OH}^-$, required moles of Mg(OH)$_2$ in $1.0 \, \text{L}$ are:

$$\frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{mol}$$

Mass in mg is:

$$5 \times 10^{-5} \times 58 \times 10^3 = 2.9 \, \text{mg}$$

Therefore, the nearest integer value is $3$. This shortcut works because the stoichiometric factor of $2$ directly connects moles of Mg(OH)$_2$ and moles of $\text{OH}^-$.