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

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Answer 1 · 2014-05-18T21:44:36

The concentration of the solution would be 0.64 mol/L.

Method 1

One way to calculate the concentration of a diluted solution is to use the formula

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

$c_{1}$ $c_1$ = 4.2 mol/L; $V_{1}$ $V_1$ = 45.0 mL = 0.0450 L
$c_{2}$ $c_2$ = ?; $V_{2}$ $V_2$ = (250 + 45.0) mL = 295 mL = 0.295 L

Solve the formula for $c_{2}$ $c_2$ .

$c_{2} = c_{1} \times \frac{V_{1}}{V_{2}}$ $c_2 = c_1 × V_1/V_2$ = 4.2 mol/L × $\frac{45.0 mL}{295 mL}$ $(45.0" mL")/(295" mL")$ = 0.64 mol/L

Method 2

Remember that the number of moles is constant

$n_{1} = n_{2}$ $n_1 = n_2$

$n_{1} = c_{1} V_{1}$ $n_1 = c_1V_1$ = 4.2 mol/L × 0.0450 L = 0.19 mol

$n_{2} = c_{2} V_{2} = n_{1}$ $n_2 = c_2V_2 = n_1$ = 0.19 mol

$c_{2} = \frac{n_{2}}{V_{2}} = \frac{0.19 mol}{0.295 L}$ $c_2 = n_2/V_2 = (0.19" mol")/(0.295" L")$ = 0.64 mol/L

This makes sense. You are increasing the volume by a factor of about 7. So the concentration should decrease by a factor of about 7.

$\frac{4.2}{7}$ $4.2/7$ = 0.6