# It is desired to prepare 4.0 M nitric acid from the available acid solution of strength 1.4 M & 6.8 M respectively. If the total volume of the 4.0 M nitric acid required to be prepared is 4 #dm^3#, calculate the volume of the 2 acid solutions to be mixed?

##### 2 Answers

You need

#### Explanation:

First we find the total moles of

So

Let volume of the 1.4M solution =

Let volume of the 6.8M solution =

So we can write:

and since the total volume =

These are called simultaneous equations. There are 2 equations and 2 unknowns which we can easily solve:

From

We can then substitute this for

So from

The volumes of the two solutions are **2.1 L** and **1.9 L**, respectively.

#### Explanation:

The idea behind this problem is that you need to use two equations, one that describes the *number of moles* needed for the target solution, and one that describes the *total volume* of the target solution.

You know that molarity is defined as moles of solute, in your case nitric acid, divided by liters of solution.

The target solution must have a concentration of **4.0 M** and a volume of **4 L** (

Now, let's label the volume of the **1.4-M** solution **6.8-M** solution

The total volume of the solution must be **4 L**, so you know that

Use the molarities and volumes of the stock solutions to find a relationship between the number of moles of nitric acid each solution adds to the total.

This is your second equation

Use equation

Use this value in equation

This means that

Since

I rounded the answer to two sig figs, despite the fact that you only gave one sig fig for the volume of the target solution.