Question

Suppose25.0gof solidNaOHis added to 1.50Lof an aqueous solution that is already2.40MinNaOH. Then water is added until the final volume is 400L. Determine the concentration of theNaOHin the resulting solution.

Short answer

The final concentration of NaOH in the resulting solution is 1.06M.

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 liter of solution and is denoted byM.

Thus, the molarity of a solution is given as:$\mathbf{M}\mathbf{=}\frac{\mathbf{No}\mathbf{.}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}\mathbf{(}\mathbf{in}\mathbf{}\mathbf{Litres}\mathbf{)}}$.

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

Final concentration= $\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{NaOH}}{\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{L}\mathbf{)}}$.

Find the number of moles of   added

Substitute the value of the given mass and molar mass of NaOH to find the number of

moles$\begin{array}{l}\mathrm{nNaOH}=\frac{Givenmass}{Molarmass}\\ \mathrm{nNaOH}=\frac{25.0\mathrm{g}}{40.0{\mathrm{gmol}}^{-1}}\\ \mathrm{nNaOH}=0.625\mathrm{mol}\end{array}$

Find the number of moles of NaOH in 2.40M solution

Substitute the known values in the above molarity formula as:

$2.40\mathrm{mol}{\mathrm{L}}^{-1}=\frac{\mathrm{MolesofNaOHsolution}}{1.5\mathrm{L}}$

Rearrange this equation to find the number of moles of NaOH in 2.40M solution as:

$\begin{array}{l}nNaOH=2.40{\mathrm{molL}}^{-1}\times 1.5\mathrm{L}\\ nNaOH=3.60\mathrm{mol}\end{array}$

Thus, the total number of moles of NaOH in the resulting solution will be as follows:

$\begin{array}{l}ntotal=(3.60+0.625)\mathrm{mol}\\ ntotal=4.23\mathrm{mol}\end{array}$

Determine the final concentration of NaOH 

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

$\begin{array}{l}c=\frac{4.23\mathrm{mol}\mathrm{NaOH}}{4.0\mathrm{L}}\\ c=1.06\mathrm{mol}{\mathrm{L}}^{-1}\\ c=1.06\mathrm{M}\end{array}$

Thus, the final concentration of NaOH in the resulting solution is 1.06M.