# What is the resultant pH of the solution.....?

## What is the resultant $p H$ of the solution prepared by taking a $10 \cdot m L$ volume of $H C l \left(a q\right)$ of $0.10 \cdot m o l \cdot {L}^{-} 1$ concentration, and diluting this to a $1 \cdot L$ volume?

Apr 21, 2017

#### Answer:

$p H = 3$

#### Explanation:

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

$\left[{H}_{3} {O}^{+}\right] = \frac{10 \times {10}^{-} 3 \cdot L \times 0.1 \cdot m o l \cdot {L}^{-} 1}{1 \cdot L}$

${10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

And thus $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

$- {\log}_{10} \left({10}^{-} 3\right) = 3. \ldots \ldots \ldots .$

And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition.