Question

A solution was prepared by mixing 50.00 mL of 0.100 M ${\mathrm{HNO}}_3$ and 100.00 mL of 0.200 M${\mathrm{HNO}}_3$ .Calculate the molarity of the final solution of nitric acid.

Short answer

Themolarity of the final solution of nitric acid is $0.167\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.$or $0.167\mathrm{M}$.

Step-by-step solution

Definition

A solution's molarity is a concept that refers to the amount of solute contained in one litre of the solution. It may be stated mathematically as follows:

Molarity$=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{liter}}$

Determination of number of HNO3 moles of in 50.00 mL.

A nitric acid solution with a concentration of $0.100\mathrm{M}$implies that there are $0.100$moles of ${\mathrm{HNO}}_3$ in one litre of the solution. As a result, the number of moles of ${\mathrm{HNO}}_3$contained in $50.00\mathrm{mL}$of solution may be determined as follows:

$\frac{\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3}{\mathrm{volume}\mathrm{in}\text{L}}=0.100\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.$
role="math" localid="1650360079376" $\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3=0.100\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.\times \mathrm{volume}\mathrm{in}\text{L}$

$=0.100\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.\times \left(\frac{1\mathrm{L}}{1000\mathrm{mL}}\right)$

$=0.005\mathrm{mol}$

As a result,there are $0.005\mathrm{mol}$of role="math" localid="1650360286605" ${\mathrm{HNO}}_3$ in $50.00\mathrm{mL}$$0.100\mathrm{M}$ nitric acid.

Determination of number of moles of HNO3 in src="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHhtbG5zOndycz0iaHR0cDovL3d3dy53aXJpcy5jb20veG1sL21hdGhtbC1leHRlbnNpb24iIGhlaWdodD0iMjAiIHdpZHRoPSI1NCIgd3JzOmJhc2VsaW5lPSIxNiI+PCEtLU1hdGhNTDogPG1hdGggeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzE5OTgvTWF0aC9NYXRoTUwiPjxtbj4xMDA8L21uPjxtbz4mI3hBMDs8L21vPjxtaT5tTDwvbWk+PC9tYXRoPi0tPjxkZWZzPjxzdHlsZSB0eXBlPSJ0ZXh0L2NzcyIvPjwvZGVmcz48dGV4dCBmb250LWZhbWlseT0iQXJpYWwiIGZvbnQtc2l6ZT0iMTYiIHRleHQtYW5jaG9yPSJtaWRkbGUiIHg9IjEzLjUiIHk9IjE2Ij4xMDA8L3RleHQ+PHRleHQgZm9udC1mYW1pbHk9IkFyaWFsIiBmb250LXNpemU9IjE2IiB0ZXh0LWFuY2hvcj0ibWlkZGxlIiB4PSI0Mi41IiB5PSIxNiI+bUw8L3RleHQ+PC9zdmc+" role="math" localid="1650360463091" 100 mL

Likewise, the number of moles of ${\mathrm{HNO}}_3$contained in $100\mathrm{mL}$of solution with the concentration of may be determined as follows:

$\frac{\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3}{\mathrm{volume}\mathrm{in}\text{L}}=0.200\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.$

$\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3=0.200\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.\times \mathrm{volume}\mathrm{in}\text{L}$

$=0.200\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.\times 100.00\mathrm{mL}\times \left(\frac{1\mathrm{L}}{1000\mathrm{mL}}\right)$

$=0.02\mathrm{mol}$

As a result,there are $0.02\mathrm{mol}$of role="math" localid="1650360861776" ${\mathrm{HNO}}_3$in $100.00\mathrm{mL}$$0.200\mathrm{M}$nitric acid.

Determination of final concentration of HNO3

The volume of the new solution, as well as the number of moles of ${\mathrm{HNO}}_3$, will be added when the two solutions are combined. As a result, the molarity of the mixed nitric acid solution is as follows:

Molarity of final ${\mathrm{HNO}}_3$solution$=\frac{\mathrm{Total}\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3}{\mathrm{Total}\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{liter}}$

$=\frac{\left(0.005+0.02\right)\mathrm{mol}}{\left(50.00\mathrm{mL}+100.00\mathrm{mL}\right)\left(\frac{1\mathrm{L}}{1000\mathrm{mL}}\right)}$

$=\frac{0.025\mathrm{mol}}{0.15\mathrm{L}}$

role="math" localid="1650361163077" $=0.167\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.$

As a result, the nitric acid solution's final concentration is $0.167\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.$or $0.167\mathrm{M}$.