# If 25*cm^3 of 1.0*mol*L^-1 nitric acid are reacted with. 50*cm^3 0.5*mol*L^-1 NaOH what is the pH?

Mar 29, 2017

$p H = 7$

#### Explanation:

We interrogate the following acid-base reaction:

$H N {O}_{3} \left(a q\right) + K O H \left(a q\right) \rightarrow K N {O}_{3} \left(a q\right) + {H}_{2} O$.

We use the old faithful formula:

$\text{Concentration}$ $\equiv$ $\text{Moles of solute"/"Volume of solution}$,

So with respect to nitric acid we have,

$25 \cdot c {m}^{3} \times {10}^{-} 3 \cdot L \cdot c {m}^{-} 3 \times 1.0 \cdot m o l \cdot {L}^{-} 1 = 0.025 \cdot m o l$.

And with respect to potassium hydroxide we have,

$50 \cdot c {m}^{3} \times {10}^{-} 3 \cdot L \cdot c {m}^{-} 3 \times 0.5 \cdot m o l \cdot {L}^{-} 1 = 0.025 \cdot m o l$

And thus there are equimolar quantities of nitric acid and sodium hydroxide, and at the end of addition we have a $75 \cdot c {m}^{3}$ volume that is stoichiometric in potassium nitrate. And since potassium nitrate does not affect solution $p H$, we have a NEUTRAL solution, $p H = 7$.