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

Answer:

pH = 13.7.

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:

$pH=14-0.30\\\\pH=13.7$

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