How many liters of 12.3 molar NaOH stock solution will be needed to make 12.4 liters of a 1.2 molar NaOH solution? Show the work used to solve this problem.
There is hardly any work involved.
mL x M = mL x M
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To solve this problem, you need to use the formula for dilution:
\(C_1V_1 = C_2V_2\)
where:
\(C_1\) = initial concentration of the stock solution
\(V_1\) = volume of the stock solution
\(C_2\) = final concentration of the diluted solution
\(V_2\) = final volume of the diluted solution
We are given:
\(C_1 = 12.3 M\)
\(C_2 = 1.2 M\)
\(V_2 = 12.4 L\)
First, we need to find \(V_1\):
\(C_1V_1 = C_2V_2\)
\(12.3 M \times V_1 = 1.2 M \times 12.4 L\)
Next, we can rearrange the equation to solve for \(V_1\):
\(V_1 = \frac{{1.2 M \times 12.4 L}}{{12.3 M}}\)
Simplifying the equation,
\(V_1 = \frac{{14.88}}{{12.3}}\)
\(V_1 = 1.21 L\)
Therefore, you would need approximately 1.21 liters of the 12.3 M NaOH stock solution to make 12.4 liters of a 1.2 M NaOH solution.