Answer:
pH = 13.7.
Explanation:
Hello there!
In this case, as we set up the chemical reaction between nitric acid and sodium hydroxide:
[tex]NaOH+HNO_3\rightarrow NaNO_3+H_2O[/tex]
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:
[tex]n_{acid}=0.25L*1.5mol/L=0.375mol\\\\n_{base}=0.25L*2.5mol/L=0.625mol[/tex]
In such a way, since the base react with more moles, there is leftover that we compute as shown below:
[tex]n_{base}^{leftover}=0.25mol[/tex]
Afterwards, we compute the concentration given the new volume of 500 mL (0.500 L), as both volumes are added up:
[tex][base]=0.25mol/0.500L=0.5M[/tex]
Now, since sodium hydroxide is such a strong base, we compute the pOH first:
[tex][OH^-]=[base]=0.5M[/tex]
[tex]pOH=-log([OH^-])=-log(0.5M)\\\\pOH=0.30[/tex]
And the pH:
[tex]pH=14-0.30\\\\pH=13.7[/tex]
Best regards!
