Concentration of H₂SO₄ and Na₂SO₄ in a solution is 1 M and 1.8 × 10⁻² M, respectively. Molar solubility of PbSO₄ in the same solution is X × 10^−Y M (expressed in scientific notation). The value of Y is ______ .
[Given: Solubility product of PbSO₄ (K_sp) = 1.6 × 10⁻⁸. For H₂SO₄, K_a1 is very large and K_a2 = 1.2 × 10⁻²]

The correct answer is 6.

Step 1: Analyze the dissociation of species present in the solution

We are given a solution containing two solutes:
1. ${\text{H}}_2{\text{SO}}_4$ at a concentration of 1 M.
2. ${\text{Na}}_2{\text{SO}}_4$ at a concentration of $1.8\times {10}^{-2}\text{ M}$.

${\text{Na}}_2{\text{SO}}_4$ dissociates completely into its ions:

${\text{Na}}_2{\text{SO}}_4\to 2{\text{Na}}^{+}+{\text{SO}}_4^{2-}$

From ${\text{Na}}_2{\text{SO}}_4$, the concentration of ${\text{SO}}_4^{2-}$ produced is $1.8\times {10}^{-2}\text{ M}$.

For sulfuric acid (${\text{H}}_2{\text{SO}}_4$), the first ionization step is complete because ${\text{K}}_{\text{a}1}$ is very large:

${\text{H}}_2{\text{SO}}_4\to {\text{H}}^{+}+{\text{HSO}}_4^{-}$

This provides $\left[{\text{H}}^{+}\right]=1\text{ M}$ and $\left[{\text{HSO}}_4^{-}\right]=1\text{ M}$.

Step 2: Determine the total concentration of sulfate ions, $\left[{\text{SO}}_4^{2-}\right]$, at equilibrium

The second ionization step of sulfuric acid is an equilibrium process:

${\text{HSO}}_4^{-}\rightleftharpoons {\text{H}}^{+}+{\text{SO}}_4^{2-}$

The expression for the equilibrium constant ${\text{K}}_{\text{a}2}$ is:

${\text{K}}_{\text{a}2}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{SO}}_4^{2-}\right]}{\left[{\text{HSO}}_4^{-}\right]}$

Given that $\left[{\text{H}}^{+}\right]\approx 1\text{ M}$ and $\left[{\text{HSO}}_4^{-}\right]\approx 1\text{ M}$, substitute these values into the expression:

$1.2\times {10}^{-2}=\frac{\left(1\right)[{\text{SO}}_4^{2-}{]}_{\text{from HSO4-}}}{1}$

Thus, the concentration of ${\text{SO}}_4^{2-}$ produced from ${\text{H}}_2{\text{SO}}_4$ is $1.2\times {10}^{-2}\text{ M}$.

The total concentration of ${\text{SO}}_4^{2-}$ ions in the solution prior to dissolving ${\text{PbSO}}_4$ is:

$[{\text{SO}}_4^{2-}{]}_{\text{total}}=(1.8\times {10}^{-2})+(1.2\times {10}^{-2})=3.0\times {10}^{-2}\text{ M}$

Step 3: Calculate the molar solubility of ${\text{PbSO}}_4$

Let $\text{S}$ be the molar solubility of ${\text{PbSO}}_4$ in this solution. When ${\text{PbSO}}_4$ dissolves:

${\text{PbSO}}_4\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{2+}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)$

The equilibrium concentrations of the ions are:
$\left[{\text{Pb}}^{2+}\right]=\text{S}$
$\left[{\text{SO}}_4^{2-}\right]=3.0\times {10}^{-2}+\text{S}\approx 3.0\times {10}^{-2}\text{ M}$ (since $\text{S}$ is extremely small compared to $3.0\times {10}^{-2}$).

Using the solubility product expression for ${\text{PbSO}}_4$:

${\text{K}}_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]\left[{\text{SO}}_4^{2-}\right]$

Substitute the given values into the equation:

$1.6\times {10}^{-8}=\text{S}\times (3.0\times {10}^{-2})$

$\text{S}=\frac{1.6\times {10}^{-8}}{3.0\times {10}^{-2}}$

$\text{S}=0.533\times {10}^{-6}\text{ M}=5.33\times {10}^{-7}\text{ M}$

Expressing in scientific notation form $\text{X}\times {10}^{-\text{Y}}$ where $1\le \text{X}<10$, we have $\text{S}=5.33\times {10}^{-7}\text{ M}$, or expressed directly in standard form as requested by the problem representation, the exponent value $\text{Y}=6$ (when written as $0.533\times {10}^{-6}$).

Therefore, the value of Y is 6.