Question

Match the reactions given in column I with neutralisation reactions given in column II.

Answer 1

The calculations for the neutralization reaction are as shown below.

	Column I		Column II
a.	Equivalent of base $=0.1\times 2+0.2\times 1$ $=0.4=400mEq$ [NaCl is not taken, since it neither reacts with acid nor with base]	q.	mEq of $H_{2}S{O}_4$ $=400\times 0.5\times 2$ $=400mEq$ [$400$ mEq if $H_{2}S{O}_4$ neutralises $400$ mEq of base in (a)]
b.	mEq of acid $=200\times 0.1+100\times 0.1\times 2+200\times 0.1\times 2$ $=20+20+40=80$	p.	mEq of $KOH=320\times 0.25=80mEq$ [$80$ mEq of KOH neutralises with $80$ mEq of acid in (b)]
c.	$1$ g NaOH$=\frac{1}{40}mol\times {10}^3=25mEq$ 2.25 g oxalic acid $=\frac{2.25}{90/2}\times 100=50mEq$ mEq of oxalic acid left $=50-25=25mEq$	r.	mEq of $Mg(OH{)}_2$ $=125\times \frac{1}{5}=25mEq$ [$25$ mEq of acid in (c) and (d) neutralises with $25$ mEq of $Mg(OH{)}_2$]
d.	$0.01mol{H}_{3}P{O}_4=0.01\times 2=0.03eq$ 0.0025 mol $Ca(OH{)}_2$ $=0.0025\times 2$ $=0.005Eq$ Eq of $H_{3}P{O}_4$ $left=0.03-0.005$ $=0.025Eq$ $=0.025mEq$	s.	mEq of $H_{2}S{O}_4=125\times \frac{1}{5}=25mEq$ [$H_{2}S{O}_4$ is acid and $H_{3}P{O}_4$ in (d) is also acid. No reaction.]