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

Answer: 0.132 M

Step-by-step 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* M_1* 100=2* 0.200* 32.9`

`M_1=0.132`

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

Answer 2

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