Question

Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.110 L0.110 L sample of an unknown HNO3HNO3 solution required 51.9 mL51.9 mL of 0.100 M Ba(OH)20.100 M Ba(OH)2 for complete neutralization. What is the concentration of the HNO3HNO3 solution? concentration:

Answer 1

Answer: The concentration of $HNO_3$ solution will be 0.094 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

Conversion factor: 1L = 1000 mL

$n_1=1\\M_1=?M\\V_1=0.110L=110mL\\n_2=2\\M_2=0.100M\\V_2=51.9mL$

Putting values in above equation, we get:

$1\times M_1\times 110=2\times 0.100\times 51.9\\\\M_1=0.094M$

Hence, the concentration of $HNO_3$ solution will be 0.094 M.