What volume of water would be added to 16.5 mL of a 0.0813 M solution of sodium borate in order to get a 0.0200 M solution?

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Answer 1 · 2014-05-05T11:19:57

To solve this we'll need to apply the equation

$M_{1} V_{1} = M_{2} V_{2}$ $M_1V_1 = M_2V_2$

$V_{1} = 16.5 m l$ $V_1 = 16.5ml$
$V_{2} = ?$ $V_2 = ?$
$M_{1} = 0.0813$ $M_1 = 0.0813$
$M_{2} = 0.200$ $M_2 = 0.200$

Solve the equation for V2
$V_{2}$ $V_2$ = $\frac{M_{1} V_{1}}{M_{2}}$ $(M_1V_1)/M_2$

$V_{2}$ $V_2$ = $\frac{0.0813 M . 16.5 m l}{0.0200 M}$ $(0.0813M . 16.5ml) / (0.0200M$ = $67.1 m l$ $67.1ml$

Please note the question asks you to find the volume that must be added. You'll need to subtract 16.5mL from 67.1 to find the answer of
50.6mL.

Here is a video which discusses how to perform dilution calculations.