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pH of a Solution

For 100 mL of...

Question

For 100 mL of a 0.3 M $C a C l_{2}$ solution + 400 mL of a 0.1 M HCl solution, which of the following is true?

A

Total concentration of cations =

0.12 M

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B

Total concentration of cations = 0.07 M

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C

Concentration of

C l^{-}

ions = 0.1 M

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D

Concentration of

C l^{-}

ions = 0.2 M

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Solution

The correct option is D Concentration of $C l^{-}$ ions = 0.2 M
We have two cations $H^{+}$ and $C a^{2 +}$ and one anion $C l^{-}$ in the resultant solution.
Let $M_{1}$ be the molarity and $V_{1}$ be the volume of $C a C l_{2}$ solution; $M_{2}$ be the molarity and $V_{2}$ be the volume of HCl solution.
Molarity of the cation = $\frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}} = \frac{(0.3 \times 100) + (0.1 \times 400)}{500} = \frac{0.7}{5} = 0.14 M$
Molarity of $C l^{-} = \frac{(2 \times M_{1}) V_{1} + M_{2} V_{2}}{V_{1} + V_{2}} = \frac{(2 \times 0.3) \times 100 + (0.1 \times 400)}{500} = \frac{0.6 + 0.4}{5} = 0.2 M$
(Since each $M g C l_{2}$ molecule has two $C l^{-}$ ions, the concentration of $C l^{-}$ in it will be $2 \times M_{1}$ )

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