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

Step-by-step 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* M_1* 105=2* 0.2* 29.1`

`M_1=0.11M[/tex<strong>] </strong></p><p><strong>Therefore, the concentration of [tex]HNO_3` will be 0.11 M.