Question

A 0.100-l sample of an unknown hno3 solution required 32.9 ml of 0.200 m ba(oh)2 for complete neutralization. what was the concentration of the hno3 solution?

Answer 1

The reaction between HNO3 and Ba(OH)2 is given by the equation below;
2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O
Moles of Barium hydroxide used;
 = 0.200 × 0.039 l
 = 0.0078 Moles
The mole ratio of HNO3 and Ba(OH)2 is 2: 1
Therefore; moles of nitric acid used will be;
= 0.0078 ×2 = 0.0156 moles
But; 0.0156 moles are equal to a volume of 0.10
The concentration of Nitric acid will be; 
 =  (0.0156 × 1)/0.1 
 = 0.156 M

Answer 2

Answer: 0.132 M

Explanation:

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 = 0.100 L = 100 ml (1L=1000ml)

$M_2$ = molarity of $Ba(OH)_2$ solution = 0.200 M

$V_2$ = volume of $Ba(OH)_2$ solution = 32.9 ml

$n_1$ = valency of $HNO_3$ = 1

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

$1\times M_1\times 100=2\times 0.200\times 32.9$

$M_1=0.132$

Therefore, the concentration of $HNO_3$ solution is 0.132 M