Question

$20\mathrm{ml}$ of $0.2\mathrm{M}$ ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ is mixed with $20\mathrm{ml}$ of $6.6\mathrm{M}$ ${\mathrm{BaCl}}_2$ , the concentration of ${\mathrm{Cl}}^{-}$ ion in solution is:

• A. $0.2$
• B. $6.6$
• C. $0.02$
• D. $0.06$

Answer 1

The correct option is B

$6.6$

Explantion for the correct option:

option c):

Step1: Given data

Molarity of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ solution = $0.2\mathrm{M}$

Volume of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$solution = $20\mathrm{ml}$

Molarity of ${\mathrm{BaCl}}_2$ solution = $6.6\mathrm{M}$

Volume of ${\mathrm{BaCl}}_2$solution = $20\mathrm{ml}$

Step 2: Formula used

The concentration of chloride ion in the solution is calculated using the formula :

= $\frac{Totalamountofcholrideinsolution}{Totalamountofsolution}$

Step 3: Calculating the total amount of chloride present in the solution

The total amount of chloride present in the solution

$$\begin{gathered} =2\times VolumeofBaC{l}_2\times MolarityofBaC{l}_2 \\ =2\times 20ml\times 6.6mmol/L \\ =264mmol \end{gathered}$$

Step 4: Calculating the total amount volume of the mixed solution

On mixing the two solutions, the total volume of the solution is calculated by adding the volumes of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$solution and ${\mathrm{BaCl}}_2$solution.

Total volume of solution = Volume of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ + Volume of ${\mathrm{BaCl}}_2$

=$20\mathrm{ml}+20\mathrm{ml}$ = $40\mathrm{ml}$

Step 5: Calculating the concentration of chloride ion in the solution

Putting the values in the formula to calculate the concentration of chloride ions in the solution.

$\frac{Totalamountofcholrideinsolution}{Totalamountofsolution}=\frac{264}{40}=6.6M$

Hence, the concentration of chloride ions in the solution is $6.6\mathrm{M}.$