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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?

User Kitcha
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2 Answers

5 votes

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

User Shaohua Huang
by
6.3k points
5 votes
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
User Sergey V
by
6.8k points