What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M KOH to 250 mL?

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Answer 1 · 2014-08-10T20:09:37

The concentration would be 0.76 mol/L.

The most common way to solve this problem is to use the formula

$c_{1} V_{1} = c_{2} V_{2}$ $c_1V_1 = c_2V_2$

In your problem,

$c_{1}$ $c_1$ = 4.2 mol/L; $V_{1}$ $V_1$ = 45.0 mL
$c_{2}$ $c_2$ = ?; $V_{2}$ $V_2$ = 250 mL

$c_{2} = c_{1} \times \frac{V_{1}}{V_{2}}$ $c_2 = c_1 × V_1/V_2$ = 4.2 mol/L × $\frac{45.0 mL}{250 mL}$ $(45.0"mL")/(250"mL")$ = 0.76 mol/L

This makes sense. You are increasing the volume by a factor of about 6, so the concentration should be about ¹/₆ of the original (¹/₆ × 4.2 = 0.7).