Question

A 25.0 mL solution of HNO3 is neutralized with 22.4 mL of 0.250 M Ba (OH)2. What is the concentration of the original HNO3 solution?

Answer 1

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

As per the reaction,

$2\times$ Moles of $Ba(OH)_2$ reacted$=$ Moles of $HNO_3$ reacted.

And,Molarity equation for the reaction will be:

$\frac{M_1V_1}{n_1}=\frac{M_2V_2}{n_2}$ ,where $M,V,n$ represent concentration,volume and stochiometric cofficients for each of the reactants and $1$ represent for $Ba(OH)_2$ and $2$ for $HNO_3.$

$\implies \frac{22.4\times0.250}{1}=\frac{25\times\ M_2}{2}$

$\implies M_2=0.448\ M=$ Concentration of the original $HNO_3$ solution