Ionization of acids and Bases

Solved Examples

Question 1
Calculate pH of solution having [H+] = $1.0 \times 10^{-3}$ M
Solution
$pH = -log [H^+]$
$pH = -log [10^{-3}]$
pH = -(-3)
pH = 3

Question 2
Give pH of cold drink having [H+] concentration $10^{-5}$
Solution
$pH = - log [H^+]= - log [10^{-5}]= -(-5)= 5$

Question 3
If NaOH solution has [OH-] = $10^{-6}$. pH = ?
Solution
$pK_w = pOH + pH$
$14 = 6 + pH$
$pH=8$

Question 4
The concentration of [H+] in the sample of cold drink is $3.8 \times 10^{-3}$. Give pH.
Solution
pH = - log [H+]
$= - log [3.8 \times 10^{-3}]$
= -[-3 + .58]
= 2.4 pH

Question 5
pH = 3.76. Give [H+]
Solution
pH = - log [H+]
3.76 = - log [H+]
$[H^+] =5.76 \times 10^{-3}$

Question 6
Calculate pH of a $1.0 \times 10^{–8}$ M solution of HCl.
Solution
$2H_2O (l) \rightleftharpoons H_3O^+(aq) + OH^– (aq)$
$Kw = [OH^-][H_3O^+]= 10^{–14}$
Let, $x = [OH^–] = [H_3O^+]$ from $H_2O$.
The $H_3O^+$ concentration is generated
(i) from the ionization of HCl dissolved i.e.,
$HCl(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + Cl^–(aq)$
(ii) from ionization of $H_2O$. In these very dilute solutions, both sources of $H_3O^+$ must be considered:
$[H_3O^+] = 10^{–8} + x$
$Kw = (10^{–8} + x)(x) = 10^{–14}$
or $x^2 + 10^{–8} x – 10^{–14} = 0$
$[OH– ] = x = 9.5 \times 10^{–8}$
So, pOH = 7.02 and pH=6.98$