# A chemist only has a stock solution of 8.0 M HCl. He needs 550 ml of a 4.5 M solution in order to complete an experiment. How would he do that?

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The chemist needs a particular volume of the stock solution in order to make the desired amount and concentration. This kind of problem uses the dilution method that can be expressed as:

`M_1 V_1 = M_2 V_2`

Where:

`M_1 = (550 mL)/(1000mL) = 0.550 L`

`V_1 = 4.5 M`

`M_2 = 8.0 M`

`V_2 = ?`

`M_1 V_1 = M_2 V_2`

`V_2 = (M_1 V_1)/M_2`

`V_2 = ((0.550)(4.5))/(8.0)`

`V_2 =0.309375 L = 310 mL`

**Amount of water to be added = 550 -310 = 240 mL**

To prepare the solution, measure 310 mL of the stock solution and 240 mL of water. Remember to add acid into water and not the other way to prevent violent chemical reactions.

**Sources:**

This is a dilution problem. The equation for diluting an equation is:

`"M"_(1)"V"_(1) = "M"_(2)"V"_(2)`

Your initial molarity is `"M"_(1)` , while your intended molarity and volume would be 2. Plug in all your values.

`(8"M")("V"_1) = (4.5"M")(550"mL")`

You can skip converting mL to L because it does not affect the answer of the problem (since you did not specify what the final units had to be). As long as the units on each side of the equation matches the other, you're good!

Solve for your unknown.

`"V"_(1) = 309.375 "mL" ~~ 310 "mL"`

You need two sig figs because all your initial values have two sig figs!