Question

What mass of magnesium hydroxide is required to neutralize $125mL$ of $0.136MHCl$ solution?

(Molar mass of $Mg(OH{)}_2=58.33gmo{l}^{-1}$)

• A. $0.248g$
• B. $0.496g$
• C. $0.992g$
• D. $1.98g$

Answer 1

Answer: (B)

$0.496g$

$mg(OH{)}_2+2HCl\to MgC{l}_2+2H_{2}O$

From the Equation, it is clear that we need

2 moles of HCl to neutralise 1 mole of $Mg(OH{)}_2$

Now there is $0.136M$ of HCl for every 1000 ml

So the sample will have,

$125ml\times \frac{0.136molesHCl}{1000ml}$

$=0.017molesHCl$

So for complete neutralisation,

$\Rightarrow 0.017molesHCl\times \frac{1moleMg\left(O{H}_2\right)}{2molesHCl}$

$\Rightarrow 0.0085molesMg(OH{)}_2$

$\Rightarrow 0.0085\times 58.33$

$=0.495$

So, $0.495g$ of $Mg(OH{)}_2$ is required to neutralise

$125ml$ of $0.136MHClSo{l}^n$.