# Question #d798c

May 6, 2017

$\text{pH} = 7$

#### Explanation:

For starters, you know that you have

${\text{HNO"_ (3(aq)) + "KOH"_ ((aq)) -> "KNO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

The two reactants react in a $1 : 1$ mole ratio, which implies that in order to have a complete neutralization, you need to add equal numbers of moles of nitric acid and of potassium hydroxide.

Now, you're adding ${\text{25 cm}}^{3}$ of $\text{1 M}$ nitric acid solution and ${\text{50 cm}}^{3}$ of $\text{0.5 M}$ potassium hydroxide solution.

Notice that the volume of the sodium hydroxide solution is double that of the nitric acid solution, but that its concentration is half that of the nitric acid solution.

This should let you know what you're adding equal numbers of moles of the two reactants.

Think of it like this. If you had ${\text{25 cm}}^{3}$ of both solutions, the nitric acid solution would contain $\textcolor{red}{2}$ times as many moles of solute because

$\text{1 M" = color(red)(2) xx "0.5 M}$

By doubling the volume of the potassium hydroxide solution, you're ensuring that the number of moles of potassium hydroxide is equal to the number of moles of nitric acid.

You can thus say that the two reactants will be completely consumed by the reaction $\to$ the resulting solution will contain water and aqueous potassium nitrate.

Therefore, the $\text{pH}$ of the solution will be equal to that of pure water at room temperature, i.e. $\text{pH} = 7$.