# Question #18150

Dec 5, 2017

$p H \approx 10.8$

#### Explanation:

I will assume that this is at 298K.

$500 m L = 0.5 L = 0.5 {\mathrm{dm}}^{3}$

$n \left(N a O H\right) = \frac{0.0125}{40} = 3.125 \cdot {10}^{- 4}$ $m o l$

$\left[N a O H\right] = \frac{3.125 \cdot {10}^{- 4}}{0.5} = 6.25 \cdot {10}^{- 4}$ $m o l$ ${\mathrm{dm}}^{- 3}$

Since NaOH is a strong base we will assume it fully dissociates.

$\left[O {H}^{-}\right] = 6.25 \cdot {10}^{- 4}$ $m o l$ ${\mathrm{dm}}^{- 3}$

$\left[{H}^{+}\right] = \frac{{K}_{w}}{\left[O {H}^{-}\right]} = \frac{1 \cdot {10}^{- 14}}{6.25 \cdot {10}^{- 4}} = 1.6 \cdot {10}^{- 11}$ $m o l$ ${\mathrm{dm}}^{- 3}$

$p H = - \log \left(\left[{H}^{+}\right]\right) = - \log \left(1.6 \cdot {10}^{- 11}\right) \approx 10.8$

Dec 5, 2017

pH$= 10.8$

#### Explanation:

First find the molarity of 0.0125g of 500mL NaOH solution

$M = \textcolor{red}{\text{Number of MOLES of solute")/color(red)("Litre}}$

$M = \frac{\frac{0.0125}{40}}{0.5 L} \text{moles" rArr "0.000625 moles per Litre}$

$\textcolor{red}{\Rightarrow} 6.25 \times {10}^{-} 4$ $\text{mol/L}$

The concentration of $O {H}^{-}$ ions $\textcolor{red}{\Rightarrow} 6.25 \times {10}^{-} 4$ $\text{mol/L}$

$p \left(O H\right) = - \log \left\{O {H}^{-} \text{ concentration}\right\}$

$p \left(O H\right) = - \log \left\{6.25 \times {10}^{-} 4\right\}$

$p \left(O H\right) = 4 - \log \left\{6.25\right\}$

$p \left(O H\right) = 4 - 0.7958$

$p \left(O H\right) = 3.2042 \Rightarrow 3.2$

Since

$p \left(O H\right) + p \left(H\right) = 14$

you have

$p H = 14 - p \left(O H\right) \Rightarrow 14 - 3.2 = 10.8$