# 100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?

## 100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?

Sep 18, 2016

We use logarithms and powers of 10 to convert between pH numbers and concentrations of ${H}^{+}$ in acids.

The pH changes from 2.6 to 2.9. It moves closer to a neutral pH of 7.

#### Explanation:

pH is defined as 'minus the logarithm (in base 10) of the concentration of ${H}^{+}$'. To find the original concentration of ${H}^{+}$, we take the pH, change its sign from positive to negative and raise 10 to that power.

That means the initial concentration of ${H}^{+}$ (sometimes written as [${H}^{+}$]) is ${10}^{- 2.6}$ = $2.51 \times {10}^{-} 3$ $m o l {L}^{-} 1$.

If we double the amount of solvent (water) and leave the amount of solute (HCl) constant, the concentration of the solution will halve, because $C = \frac{n}{V}$, concentration equals number of moles of solute divided by volume of solution.

The new concentration of ${H}^{+}$ will be half of $2.51 \times {10}^{-} 3$ = $1.26 \times {10}^{-} 3$ $m o l {L}^{-} 1$.

To find the pH of this final solution, we reverse the process. We take the logarithm (to base 10):

${\log}_{10} \left(1.26 \times {10}^{-} 3\right)$ = -2.90

Then we change the sign back to positive, and the new pH is 2.90.

(Note: because HCl is a 'strong acid', I have assumed throughout that it is completely dissociated in solution)

Sep 18, 2016

$2.901$

#### Explanation:

Here's a quick way of solving this problem. You know that the pH of the solution is given by the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$.

When you double the volume of the solution, you essentially halve the concentration of hydronium cations. This means that if you take ${\left[{\text{H"_3"O}}^{+}\right]}_{0}$ to be initial concentration of hydronium cations, you will have

["H"_ 3"O"^(+)]_ "dil" = (["H"_ 3"O"^(+)]_0)/2 -> the concentration after the dilution

You know that

color(purple)(bar(ul(|color(white)(a/a)color(black)("pH" = - log(["H"_3"O"^(+)]))color(white)(a/a)|)))

You can thus say that after the solution is diluted, its pH will be

"pH"_ "dil" = - log(["H"_ 3"O"^(+)]_"dil")

This, in turn, is equivalent to

"pH"_ "dil" = - log((["H"_ 3"O"^(+)]_0)/2)

"pH"_ "dil" = -[log(["H"_ 3"O"^(+)]_ 0) - log2]

"pH"_ "dil" = log2 - log(["H"_3"O"^(+)]_0)

But since

"pH"_ 0 = - log(["H"_ 3"O"^(+)]_0) = 2.600

you will have

${\text{pH"_ "dil" = log2 + "pH}}_{0}$

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH" _ "dil} = 2.600 + \log 2 = 2.901} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three decimal places.

Finally, does this result make sense?

Diluting the solution means decreasing its concentration of hydronium cations, which in turn implies increasing the pH of the solution, i.e. making it less acidic.