# How can you find the pH if 75 grams of HCl (hydrochloric acid) is put in 5.0L of water?

## I understand that I have to convert the grams to mol, and I have the molarity, but I don't know if the concentration is of hydrogen or hydroxide ions. Can anybody help me? Thanks!

May 11, 2018

#### Answer:

Well, let's see...I gets $p H = 0.386$

#### Explanation:

We assume that hydrogen chloride undergoes complete protonolysis in aqueous solution...

$H C l \left(g\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + C {l}^{-}$

And so we calculate the concentration of $H C l$ in the usual way...

$\left[H C l\right] = \text{Moles of HCl"/"Volume of solution}$

We ASSUME (reasonably) that the volume of THE SOLUTION is $5.0 \cdot L$...it would be very close...

And so...$\left[H C l\right] = \frac{\frac{75 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}}{5.0 \cdot L} = 0.411 \cdot m o l \cdot {L}^{-} 1$

And so it is $0.411 \cdot m o l \cdot {L}^{-} 1$ with respect to hydronium ions... which we represent as ${H}_{3} {O}^{+}$ or ${H}^{+}$...

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(0.411\right) = - \left(- 0.386\right) = + 0.386$

The maximum concentration of $H C l$ is approx. $10.6 \cdot m o l \cdot {L}^{-} 1$...what is $p H$ here...?