# Question b50ac

Jul 10, 2017

You can add $900$ $\text{mL}$ of distilled water.

#### Explanation:

We're asked to find out how to dilute a $100$-$\text{mL}$ $0.1 M$ $\text{HCl}$ solution to a concentration of $0.01 M$.

We can dilute solutions like this by adding a certain amount of distilled water until it reaches the desired concentration. We therefore need to find out what the final volume of the solution must be to have a concentrationof $0.01 M$.

We can use the dilution equation

${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

where

• ${M}_{1}$ and ${M}_{2}$ are the molarities of the concentrated and dilute solutions, respectively

• ${V}_{1}$ and ${V}_{2}$ are the volumes of the concentrated and dilute solutions, respectively

We know:

${M}_{1} = 0.1 M$

${V}_{1} = 100$ $\text{mL}$

${M}_{2} = 0.01 M$

Plugging these in and solving for ${V}_{2}$:

${V}_{2} = \frac{{M}_{1} {V}_{1}}{{M}_{2}} = \frac{\left(0.1 \cancel{M}\right) \left(100 \textcolor{w h i t e}{l} \text{mL}\right)}{0.01 \cancel{M}} = {10}^{3}$ $\text{mL}$

Now that we know the necessary plume to attain a concentration of $0.01 M$, we subtract the original volume from this to find out how much distilled water must be added:

${10}^{3}$ $\text{mL}$ $- 100$ $\text{mL}$ = color(blue)(900 color(blue)("mL"#