Q3.139CP - Chemistry: Molecular Nature Of Matter And Change

Question

To 1.35 L of 0.325 M HCl, you add 3.57 L of a second HCl solution of unknown concentration. The resulting solution is 0.893 M HCl. Assuming the volumes are additive, calculate the molarity of the second HCl solution.

Step-by-Step Solution

Answer

The molarity of second HCl solution is$\text{1.11M}$

Step 1: Calculation of total volume:

Given:

 Volume of HCl solution(V₁) = $\text{1.35 L}$

Molarity of HCl solution(M₁)= $\text{0.325M}$

Volume of HCl solution(V₂) = $\text{3.57 L}$

Molarity of HCl solution(M₂)= ?

Molarity of resultant solution(M)=$\text{0.893M}$

 Total volume of resultant solution(V) = V₁ + V₂

$$\begin{gathered} \text{ = 1.35 L}+\text{3.57 L} \\ =4.92\text{L} \end{gathered}$$

Step 2: Calculation of molarity of second solution.

Let us calculate the molarity of second solution by using formula of mixing

${\text{M}}_1{\text{V}}_1+{\text{M}}_2{\text{V}}_2=\text{M}\left({\text{V}}_1+{\text{V}}_2\right)$

Substitute the values of molarity and volume in formula:

$$\begin{gathered} {\text{0.325M×1.35 L+M}}_{\text{2}}\text{×3.57 L = } \text{0.893M ×4.92L} \\ {\text{0.43875 + 3.57 L× M}}_{\text{2}}\text{ = } \text{4.39356} \\ {\text{3.57 L×M}}_{\text{2}}\text{ = } \text{4.39356 - 0.43875} \\ {\text{M}}_{\text{2}}\text{ = }\frac{\text{3.95481}}{\text{3.57}}\text{ = 1.11M} \end{gathered}$$

Result:

The molarity of second solution is $1.11 \text{M}$