Question

250 mL of 1.5 M nitric acid is mixed with 250 mL of 2.5 M sodium hydroxide. Calculate the pH of the resulting mixture.

Answer 1

pH = 13.7.

Step-by-step explanation:

Hello there!

In this case, as we set up the chemical reaction between nitric acid and sodium hydroxide:

`NaOH+HNO_3\rightarrow NaNO_3+H_2O`

It is possible to realize there is a 1:1 mole ratio of acid to base, thus, we next compute the moles of each one:

`n_(acid)=0.25L*1.5mol/L=0.375mol\\\\n_(base)=0.25L*2.5mol/L=0.625mol`

In such a way, since the base react with more moles, there is leftover that we compute as shown below:

`n_(base)^(leftover)=0.25mol`

Afterwards, we compute the concentration given the new volume of 500 mL (0.500 L), as both volumes are added up:

`[base]=0.25mol/0.500L=0.5M`

Now, since sodium hydroxide is such a strong base, we compute the pOH first:

`[OH^-]=[base]=0.5M`

`pOH=-log([OH^-])=-log(0.5M)\\\\pOH=0.30`

And the pH:

`pH=14-0.30\\\\pH=13.7`

Best regards!