Question 717a6

May 28, 2017

You must use 9.9 mL of the $\text{NaOH}$.

Explanation:

You can use the dilution formula

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {c}_{1} {V}_{1} = {c}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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#

${c}_{1} = \text{19 mol/L"; color(white)(ll)V_1 = "?}$
${c}_{2} = \text{0.15 mol/L"; V_2 = "1.25 L}$
${V}_{1} = \text{1.25 L" × (0.15 color(red)(cancel(color(black)("mol/L"))))/(19 color(red)(cancel(color(black)("mol/L")))) = "0.0099 L" = "9.9 mL}$