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 $PbSO _4\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $H _2 SO _4, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

The solubility product of $PbI _{2}$ is $8.0 \times 10^{-9} .$ The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $x \times 10^{-6} \,mol / L$. The value of $x$ is ……. .(Rounded off to the nearest integer)

$[$ Given $: \sqrt{2}=1.41]$

The concentration of $500$ $ML$ $NaOH$ solution is $0.02$ $M$. How many grams of $FeS{O_4}$ added in this solution for precipitation of $Fe{\left( {OH} \right)_2}$ ? The ${K_{sp}}$ of $Fe{\left( {OH} \right)_2}$ is $1.5 \times {10^{ – 15}}$. Molecular mass of $Fe{\left( {OH} \right)_2}$ is $ 152$ $g$ $mol^{-1}$ )

At $25\,^oC,$ the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}.$ At which $pH,$ will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001 \,M \,Mg^{2+}$ ions?

The $pH$ at which $\mathrm{Mg}(\mathrm{OH})_2\left[\mathrm{~K}_{\mathrm{sp}}=1 \times 10^{-11}\right]$ begins to precipitate from a solution containing $0.10 \ \mathrm{M}$ $\mathrm{Mg}^{2+}$ ions is_____