Concentration of H$$_2$$SO$$_4$$ and Na$$_2$$SO$$_4$$ in a solution is 1 M and 1.8 $$\times$$ 10$$^{-2}$$ M, respectively. Molar solubility of PbSO$$_4$$ in the same solution is X $$\times$$ 10$$^{-Y}$$ M (expressed in scientific notation). The value of Y is _______.

[Given: Solubility product of PbSO$$_4$$ ($$K_{sp}$$) = 1.6 $$\times$$ 10$$^{-8}$$. For H$$_2$$SO$$_4$$, $$K_{a1}$$ is very large and $$K_{a2}$$ = 1.2 $$\times$$ 10$$^{-2}$$]

First note the equilibria already present in the solution.

1. Complete dissociation of sodium sulphate:
$$Na_2SO_4 \rightarrow 2\,Na^+ + SO_4^{2-}$$
Initial $$[SO_4^{2-}] = 1.8 \times 10^{-2}\,{\rm M}$$.

2. First dissociation of sulphuric acid is virtually complete:
$$H_2SO_4 \rightarrow H^+ + HSO_4^-$$
Hence after this step
$$[H^+] = 1\;{\rm M},\qquad [HSO_4^-] = 1\;{\rm M}$$.

3. The second dissociation of $$H_2SO_4$$ is governed by $$K_{a2}=1.2\times10^{-2}:$$
$$HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$$.

Let a change of $$y$$ M occur in the above equilibrium, counted positive when the reaction proceeds to the right and negative when it goes to the left.

At equilibrium:
$$[H^+] = 1 + y$$
$$[HSO_4^-] = 1 - y$$
$$[SO_4^{2-}] = 1.8\times10^{-2} + y$$.

Apply the expression for $$K_{a2}:$$
$$K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} = 1.2\times10^{-2}$$.

Substituting the concentrations gives
$$\frac{(1+y)\,(1.8\times10^{-2}+y)}{1-y}=1.2\times10^{-2}\;-(1)$$.

Because $$y$$ is expected to be small compared with 1 we solve (1) exactly (quadratic in $$y$$) to avoid error:
$$(1+y)(0.018+y)=0.012(1-y)$$
$$0.018 + 1.018y + y^2 = 0.012 - 0.012y$$
$$y^2 + 1.03y + 0.006 = 0$$.

The positive root is not physically allowed; the negative root is
$$y = -5.85\times10^{-3}\;{\rm M}$$ (still within the limit $$y \gt -0.018$$, so valid).

Therefore the equilibrium sulphate ion concentration is
$$[SO_4^{2-}] = 0.018 + y = 0.018 - 0.00585 \approx 0.012\,{\rm M}$$.

Now let the molar solubility of $$PbSO_4$$ in this medium be $$s$$ M:
$$PbSO_4(s) \rightleftharpoons Pb^{2+} + SO_4^{2-}$$
At equilibrium:
$$[Pb^{2+}] = s,\qquad [SO_4^{2-}] = 0.012 + s \approx 0.012$$ (because $$s$$ will be very small).

Using $$K_{sp}(PbSO_4)=1.6\times10^{-8}$$:
$$K_{sp} = [Pb^{2+}][SO_4^{2-}] \approx s \times 0.012$$
$$\therefore s = \frac{1.6\times10^{-8}}{0.012} = 1.33\times10^{-6}\,{\rm M}$$.

Written in scientific notation, the solubility is $$1.33 \times 10^{-6}$$ M, i.e. $$X\times10^{-Y}$$ with $$Y = 6$$.

Hence, the required value of $$Y$$ is 6.