Question

Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.105 L sample of an unknown HNO3 solution required 29.1 mL of 0.200 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3 solution?

Answer 1

Answer: 0.11 M

Explanation:

The equation for the reaction is given as:

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HNO_3$ solution = ?

$V_1$ = volume of $HNO_3$ solution = 0.105 L= 105 ml

$M_2$ = molarity of $Ba(OH)_2$ solution = 0.2 M

$V_2$ = volume of $Ba(OH)_2$ solution = 29.1 ml

$n_1$ = valency of $HNO_3$ = 1

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

$1\times M_1\times 105=2\times 0.2\times 29.1$

$M_1=0.11M[/tex]

Therefore, the concentration of [tex]HNO_3$ will be 0.11 M.