# How do you calculate the pH of a .40 solution of Ba(OH)_2 when 25.0 mL is added to 250 mL of water?

Jul 20, 2017

Well, $p H = 12 - 13.$

#### Explanation:

We first calculate $\left[B a {\left(O H\right)}_{2}\right] = \text{Moles of stuff"/"Volume of solution}$

$= \frac{0.40 \cdot m o l \cdot {L}^{-} 1 \times 25.0 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1}{275.0 \times {10}^{-} 3 \cdot L}$

$= 0.0364 \cdot m o l$ with respect to $\left[B a {\left(O H\right)}_{2}\right]$, i.e. $0.07272 \cdot m o l \cdot {L}^{-} 1$ with respect to $\left[H {O}^{-}\right]$.

$p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(0.07272\right) = 1.14$

$p H = 14 - p O H = 14 - 1.14 = 12 - 13$.

Note that you might have been a bit careless with the wording of your question. You get a different answer when the solution is diluted to $250 \cdot m L$ rather than when $25 \cdot m L$ is added to $250 \cdot m L$ water......