Question 15.14: Calculate the pH of: (a) a 0.10 M solution of AlCl3; Ka for ......

(a) 

(Steps 1–4. The species present initially are $Al(H_{2}O)_{6}  ^{3+}$ (acid), $Cl^{-}$ (inert), and $H_{2}O$ (acid or base). Because $Al(H_{2}O)_{6}  ^{3+}$ is a much stronger acid than water $(K_{a}  >>  K_{w})$, the principal reaction is dissociation of $Al(H_{2}O)_{6}  ^{3+}$:

Table 1

Step 5. The value of x is obtained from the equilibrium equation:

Step 6. pH $= – log  (1.2\times 10^{-3}) = 2.92$

Thus, $Al(H_{2}O)_{6}  ^{3+}$ is a much stronger acid than $NH_{4}  ^{+}$, which agrees with the colors of the indicator in Figure 15.9.

(b)

Step 1. The species present initially are $Na^{+}$ (inert), $CN^{-}$ (base), and $H_{2}O$ (acid or base).

Step 2. There are two possible proton-transfer reactions:

Step 3. $K_{b}  =  K_{w}/(K_{a}  for  HCN) = 2.0\times 10^{-5}$. Because $K_{b}  >>  K_{w},  CN^{-}$ is a stronger base than $H_{2}O$ and the principal reaction is proton transfer from $H_{2}O  to  CN^{-}$.

Step 4.

Table 2

Step 5. The value of x is obtained from the equilibrium equation:

Steps 6–7. $[H_{3}O^{+}]=\frac{K_{w}}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{1.4\times 10^{-3}}=7.1\times 10^{-12}$

Step 8. pH $= – log(7.1\times 10^{-12}) = 11.15$

The solution is basic, which agrees with the color of the indicator in Figure 15.9.