The solubility of $PbC{l_2}$ at ${25\,^o}C$ is $6.3 \times {10^{ – 3}}$ $mole/litre$. Its solubility product at that temperature is

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 $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

How many grams of $Ca{C_2}{O_4}$ (molecular weight $= 128$) on dissolving in distilled water will give a saturated solution $[{K_{sp}}(Ca{C_2}{O_4}) = 2.5 \times {10^{ – 9}}$ $mo{l^2}{l^{ – 2}}]$

The minimum volume of water required to dissolve $0.1\,g$ lead $(II)$ chloride to get a saturated solution ($K_{SP}$ of $PbCl_2 = 3.2 \times 10^{-8}$; atomic mass of $Pb= 207\, u$) is……$L$

At $25\,^oC$, the solubility product of $Mg(OH)_2$ is $1.0\times10^{-11}$. 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?