Question

Calculate [OH^-], pOH, and pH for each of the following.

a. 0.00040 M Ca(OH)₂

b. a solution containing 25 g of KOH per liter

c. a solution containing 150.0 g of NaOH per liter

Short answer

pH for different solutions is given as

(a) For 0.00040 M Ca(OH)₂,

(b) For 25 g of KOH per liter,

(c) For 150.0 g of NaOH per liter,

Step-by-step solution

pH of 0.00040 M Ca(OH)2 

Ca(OH)₂ is a strong base so it dissociates almost completely. So,

2×Ca(OH)₂ = [OH^-] = 2×0.00040= 0.00080 M

$\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[0.00080\right]=3.097$

At 25°C,

$$\begin{gathered} \mathrm{pH}+\mathrm{pOH}=14 \\ \mathrm{pOH}=14-\mathrm{pH}=14-3.097=10.903 \end{gathered}$$

pH of 25 g of KOH per liter

KOH is a strong base so it dissociates completely.

$\mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar}\mathrm{mass}}=\frac{25}{56}=0.446\mathrm{Mole}$

KOH = [OH^-] =0.446 mole/L

$$\begin{gathered} \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[0.446\right]=0.35 \\ \mathrm{pH}=14-\mathrm{pOH}=14-0.35=13.65 \end{gathered}$$

pH 150.0 g of NaOH per liter

NaOH is a strong base so it will also dissociate completely.

$\mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar}\mathrm{mass}}=\frac{150}{40}=3.750\mathrm{Mole}$

NaOH = [OH^-] =3.75 mole/L

$$\begin{gathered} \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left[3.75\right]=0.574 \\ \mathrm{pH}=14-\mathrm{pOH}=14-0.574=14.574 \end{gathered}$$