# Question #77396

Jun 3, 2017

$\textsf{\left(a\right)}$

4.4

$\textsf{\left(b\right)}$

4.9

$\textsf{\left(c\right)}$

$\textsf{7.9 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{\left(a\right)}$

$\textsf{p H = - \log \left[{H}^{+}\right]}$

At this low concentration I thought that you may need to account for the $\textsf{{H}^{+}}$ ions which arise from the dissociation of water. I ran it through a calculation and found that this has negligible effect.

500 ml = 0.5 L

$\textsf{\left[{H}^{+}\right] = \frac{c}{v} = \frac{2.0 \times {10}^{- 5}}{0.5} = 4 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{p H = - \log \left(4 \times {10}^{- 5}\right) = 4.4}$

$\textsf{\left(b\right)}$

$\textsf{\left[{H}^{+}\right] = \frac{2.0 \times {10}^{- 5}}{1.5} = 1.3 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{p H = - \log \left(1.3 \times {10}^{- 5}\right) = 4.9}$

$\textsf{\left(c\right)}$

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p O H = 14 - p H = 14 - 4.9 = 9.1}$

$\textsf{\left[O {H}^{-}\right] = {10}^{- 9.1} = 7.9 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$