Question

If it requires 18.2 milliliters of 0.45 molar barium hydroxide to neutralize 38.5 milliliters of nitric acid, solve for the molarity of the nitric acid solution. Show all of the work used to solve this problem. Unbalanced equation: Ba(OH)2 + HNO3 yields Ba(NO3)2 + H2O

Answer 1

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Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O

(18.2 mL) x (0.45 M Ba(OH)2) x (2 mol HNO3 / 1 mol Ba(OH)2) / (38.5 mL HNO3) = 0.43 M HNO3

Answer 2

Answer: 0.42 M

Explanation:

$Ba(OH)_2+2HNO_3\rightarrow Ba(NO_3)_2+2H_2O$

The expression used will be :

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of $HNO_3$ = 1

$n_2$ = acidity of $Ba(OH)_2$ = 2

$M_1$ = concentration of $HNO_3$ = ?

$M_2$ = concentration of $Ba(OH)_2$ = 0.45 M

$V_1$ = volume of $HNO_3$ = 38.5 ml

$V_2$ = volume of $Ba(OH)_2$ = 18.2 ml

Now put all the given values in the above law, we get the concentration:

$(1\times M_1\times 38.5)=(2\times 0.45M\times 18.2ml)$

By solving the terms, we get :

$M_1=0.42M$

Therefore, the concentration of $HNO_3$ is 0.42M