Question

A colourless crystalline salt ${}^{\prime }{X}^{\prime }$ is soluble in dilute $HCl$. On adding $NaOH$ solution, it gives a white precipitate which is insoluble in excess of $NaOH$. ${}^{\prime }{X}^{\prime }$ is :

• A. ${Al}_2{\left(S{O}_4\right)}_3$
• B. $ZnS{O}_4$
• C. $MgS{O}_4$
• D. $Sn{Cl}_2$

Answer 1

Answer: (C)

$MgS{O}_4$

$MgS{O}_4$ is soluble in HCl and gives $MgC{l}_2+H_{2}S{O}_4\left(aq\right)$
$MgS{O}_4\left(s\right)+2HCl\left(aq\right)\to MgC{l}_2+H_{2}S{O}_4\left(aq\right)$
$M{g}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)+N{a}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to Mg\left(OH{)}_2\right(s)+N{a}^{+}(aq)+C{l}^{-}(aq)$
Then, find the net ionic equation by cancelling out any species that appear on both sides of the equation.
$M{g}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\to Mg\left(OH{)}_2\right(s)$
A colourless crystalline salt X is soluble in dilute HCl On adding $NaOH$ solution, it gives a white precipitate ($Mg(OH{)}_2$) which is insoluble in excess of $NaOH$. X is $MgS{O}_4$
Hence option C is correct.