Respuesta :
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
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: 0.132 M
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = ?
[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] solution = 0.100 L = 100 ml (1L=1000ml)
[tex]M_2[/tex] = molarity of [tex]Ba(OH)_2[/tex] solution = 0.200 M
[tex]V_2[/tex] = volume of [tex]Ba(OH)_2[/tex] solution = 32.9 ml
[tex]n_1[/tex] = valency of [tex]HNO_3[/tex] = 1
[tex]n_2[/tex] = valency of [tex]Ba(OH)_2[/tex] = 2
[tex]1\times M_1\times 100=2\times 0.200\times 32.9[/tex]
[tex]M_1=0.132[/tex]
Therefore, the concentration of [tex]HNO_3[/tex] solution is 0.132 M
