"20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is _________ $\times 10^{-2} \mathrm{M}$. (Nearest integer)"

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Accepted Answer

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To determine the final concentration of the NaOH solution, we use the formula for mixing solutions :

$ M_F = \frac{M_1 \times V_1 + M_2 \times V_2}{V_1 + V_2} $

Where :

$ M_1 = 2 \, \text{M} $ and $ V_1 = 20 \, \text{mL} $: Concentration and volume of the first solution.

$ M_2 = 0.5 \, \text{M} $ and $ V_2 = 400 \, \text{mL} $: Concentration and volume of the second solution.

Substitute the values into the equation:

$ M_F = \frac{2 \times 20 + 0.5 \times 400}{420} $

Calculating each term:

$ 2 \times 20 = 40 $

$ 0.5 \times 400 = 200 $

Add these results:

$ M_F = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.571 \, \text{M} $

Convert this to scientific notation as the problem specifies the answer should be in $ \times 10^{-2} $ form:

$ M_F = 57.1 \times 10^{-2} \, \text{M} $

Rounded to the nearest integer, the final concentration is:

$ 57 $

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20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is _________ $\times 10^{-2} \mathrm{M}$. (Nearest integer)

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20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution. The final concentration of the solution is _________ $\times 10^{-2} \mathrm{M}$. (Nearest integer)

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