Question

Suppose0.400Lof a solution of0.700Mnitric acid is added to 0.800Lof a solution of0.300Mnitric acid, giving a total volume of 1.200L. Calculate the concentration (molarity) of nitric acid in the resulting solution.

Short answer

The concentration of nitric acid in the resulting solution will be 0.0433M.

Step-by-step solution

The concept of molarity and the formula used to determine the final concentration

Molarity is defined as no. of moles of solute present in one litre of solution and is denoted by M.

Thus, the molarity of a solution is given as: $\mathit{M}\mathbf{=}\frac{\mathbf{No}\mathbf{.}\mathbf{ofmolesofsolute}}{\mathbf{Volumeofsolution}\mathbf{\left(}\mathbf{inLitres}\mathbf{\right)}}$.

The formula used to determine the final concentration is given as follows:

Final concentration$\mathbf{=}\frac{\mathbf{NumberofMolesofHNO}\mathbf{3}}{\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{200}\mathbf{L}\mathbf{)}}$.

Find the initial number of moles of nitric acid

Substitute the known values in the above molarity formula as:

$0.0700\mathrm{mol}{\mathrm{L}}^{-1}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}{\mathrm{HNO}}_3}{0.400\mathrm{L}}$

Rearrange this equation to find the initial number of moles of ${\mathrm{HNO}}_3$:

$\begin{array}{l}\mathrm{ninitial}=0.0700\mathrm{mol}{L}^{-1}\times 0.400\mathrm{L}\\ \mathrm{ninitial}=0.0280\mathrm{mol}\end{array}$

Find the final number of moles of nitric acid

Substitute the known values in the above molarity formula as:

$0.0300{\mathrm{molL}}^{-1}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}{\mathrm{HNO}}_3}{0.800\mathrm{L}}$

Rearrange this equation to find the final number of moles of ${\mathrm{HNO}}_3$

$\begin{array}{l}nfinal=0.0300{\mathrm{molL}}^{-1}\times 0.800\mathrm{L}\\ nfinal=0.0240\mathrm{mol}\end{array}$

Find the total number of moles of nitric acid 

Add the values of initial and final number of moles of ${\mathrm{HNO}}_3$ to find the total moles as:

$\begin{array}{l}ntotal=ninitial+nfinal\\ ntotal=(0.0280+0.0240)\mathrm{mol}\\ ntotal=0.0520\mathrm{mol}\end{array}$

Determine the final concentration of nitric acid

Substitute the values in the above equation to calculate the final concentration as:

$\begin{array}{l}c=\frac{0.0520\mathrm{mol}{\mathrm{HNO}}_3}{1.200\mathrm{L}}\\ c=0.0433\mathrm{mol}{\mathrm{L}}^{-1}\\ c=0.0433\mathrm{M}\end{array}$

Thus, the concentration of nitric acid in the resulting solution will be 0.0433M.