Question #717a6

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Answer 1 · 2017-05-28T21:47:35

You must use 9.9 mL of the $NaOH$ $"NaOH"$ .

You can use the dilution formula

$color(blue)(bar(ul(|color(white)(a/a)c_1V_1 = c_2V_2color(white)(a/a)|)))" "$

where $c_1$ and $c_2$ are the concentrations and $V_1$ and $V_2$ are the volumes.

You can rearrange this formula to get

$V_1 = V_2 × c_2/c_1$

Your data are:

$c_1 = "19 mol/L"; color(white)(ll)V_1 = "?"$
$c_2 = "0.15 mol/L"; V_2 = "1.25 L"$

$V_1 = "1.25 L" × (0.15 color(red)(cancel(color(black)("mol/L"))))/(19 color(red)(cancel(color(black)("mol/L")))) = "0.0099 L" = "9.9 mL"$