Calculate the molarity of the resultant solution in each of the following cases:
i. 200 mL of 1 M HCl + 300 mL water
ii. 200 mL of 1 M HCl + 100 mL of 0.5 M HCl

0.4 M, \frac{5}{6} M

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0.4 M, 0.5 M

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0.2 M, \frac{5}{6} M

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0.5 M, 0.5 M

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Solution

The correct option is A $0.4 M, \frac{5}{6} M$
(i) $M_{R} \times V_{R} = M_{1} \times V_{1}$
Where $M_{R}$ is the molarity and $V_{R}$ is the volume of the resultant solution and $M_{1}$ and $V_{1}$ is the molarity and volume of the HCl solution, respectively.
Molarity of the resultant solution, $M_{R} = \frac{M_{1} \times V_{1}}{V_{R}} = \frac{1 \times 200}{500} = 0.4 M$
(Where $V_{R}$ is the volume of the resultant solution = (200 + 300) = 500 mL)
(ii). $M_{1} V_{1} + M_{2} V_{2} = M_{R} V_{R}$
Volume of the resultant solution, $V_{R} = V_{1} + V_{2}$
Molarity of the resultant solution, $M_{R} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{R}} = \frac{(1 \times 200) + (0.5 \times 100)}{300} = \frac{5}{6} M$

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