Question

Assuming complete reaction, what volume of 0.200 mol dm3 hcl(aq) is required to neutralize 25.0 cm3 of 0.200 mol dm3 ba(oh)2(aq)?

Answer 1

Answer: $0.05dm^3$

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HCl$ solution = $0.200mol/dm^3$

$V_1$ = volume of $HCl$ solution = ?

$M_2$ = molarity of $Ba(OH)_2$ solution = $0.200mol/dm^3$

$V_2$ = volume of $Ba(OH)_2$ solution = $25.0cm^3=0.025dm^3$     $1dm^3=1000cm^3$

$n_1$ = valency of $HCl$ = 1

$n_2$ = valency of $Ba(OH)_2$ = 2

$1\times 0.200M\times V_1=2\times 0.200\times 0.025dm^3$

$V_1=0.05dm^3$

Therefore, the volume of  $HCl$ solution required to neutralize $25cm^3$ of $0.200mol/dm^3$ of $Ba(OH)_2$  is $0.05dm^3$.