Question

Question: Determine [OH^-], [H⁺], and the pH of each of the following solutions.

a. 1.0 M KCI

b. 1.0 M KF

Short answer

Answer

pH for KCl (neutral salt): 7, $\left[H^{+}\right]={10}^{-7}{\text{, [OH}}^{\text{-}}]={10}^{-7}$

pH for KF = 8.57, $\left[H^{+}\right]=2.7\times {10}^{-9},\left[O{H}^{-}\right]=3.7\times {10}^{-6}$

Step-by-step solution

KCl Dissociation

KCl is a salt of strong acid HCl and strong base KOH.

pH for KCl (neutral salt): 7

$$\begin{gathered} \left[H^{+}\right]={10}^{-pH}={10}^{-7} \\ {\text{[OH}}^{\text{-}}]=\frac{{\text{K}}_{\text{w}}}{\left[H^{+}\right]}=\frac{1\times {10}^{-14}}{{10}^{-7}}={10}^{-7} \end{gathered}$$

The H₃O⁺ ion so produce, make the solution acidic.

KF Dissociation

KF is the salt of weak acid(HF) and strong base (KOH). Its shows anionic hydrolysis like this :

$$\begin{gathered} {\text{F}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{HF + OH}}^{\text{-}} \\ \text{1}\text{0}\text{0} \\ \text{1 - x}\text{x}\text{x} \\ {\text{K}}_{\text{b}}=\frac{{\text{x}}^2}{1-\text{x}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}=\frac{1\times {10}^{-14}}{7.2\times {10}^{-4}}=1.39\times {10}^{-11}(1-\text{x}\approx 1) \\ {\text{x}}^2=1.39\times {10}^{-11}\Rightarrow \text{x}=3.7\times {10}^{-6} \end{gathered}$$

[OH^-] = 3.7×10^-6