 # pH and Solutions

Table of Content:

## pH and Solutions

Water itself ionizes and has a hydrogen concentration of 10-7 moles per litre. So, for pH near 6-8, the hydrogen ion concentration of water also need be included.

### Mixture of Two Strong Acids

The strong acid is completely ionized. So the concentration of the hydrogen ion is the same as the acid concentration. The concentration of the hydrogen ion in the mixture is the sum of the acid concentration divided by the total volume.

Say, N1, V1 are the strength and volume of the first strong acid and N2, V2 of the second acid.

The concentration of the hydrogen ion in acid 1 is N1V1 and in acid 2 is N2V2

Total hydrogen concentration = N1V1 + N2V2

Total volume of solution = V1 + V2

[H+] = $=\frac{N1V1+N2V2}{V1+V2}$ from which pH can be calculated

### Mixture of Two Strong Bases

Strong bases are completely ionized. So the concentration of the hydroxide ion is the same as the base concentration. The concentration of the hydroxide ion in the mixture is the sum of the base concentration divided by the total volume.

Say, N1, V1 is the strength and volume of the first strong base and N2, V2 of the second base.

The concentration of the hydroxide ion in base 1 is N1V1 and in base 2 is N2V2

Total hydroxide concentration = N1V1 + N2V2

The total volume of solution = V1 +V2

[OH] = $=\frac{N1V1+N2V2}{V1+V2}$ [H+] = $=\frac{10^{-14}}{[OH^{-}]}$ from which pH can be calculated

### Mixture of a Strong Acid and a Strong Base

On mixing neutralization takes place. The resulting solution may be an acid or base depending on the Concentration.

Say, N1, V1 are the strength and volume of the strong acid and N2, V2 of the base.

• If, N1V1> N2V2, resulting solution will be acidic, with [H+] = $=\frac{N1V1-N2V2}{V1+V2}$
•  If, N1V1˂ N2V2, resulting solution will be basic, with [OH] = $=\frac{N2V2-N1V1}{V1+V2}$

### Weak Acid

Weak ionize partly, and Ostwald’s dilution Law can be applied to calculate pH.

$HA\rightleftharpoons H^{+}+A^{-}$

Initial concentration, moles/l, C 0 0

At equilibrium, moles /l C(1-α) Cα Cα

So, Acid Ionization constant = $Ka=\frac{[H+][A-]}{HA}=\frac{(C\alpha +C\alpha)}{c(1-\alpha)}=\frac{c\alpha^{2}}{(1-\alpha)}$

(i) For very weak electrolytes, since α <<< 1, (1 – α ) = 1

.·. $Ka=C\alpha^{2\;\;\;\;\;}\alpha = \sqrt{\frac{Ka}{c}}=\sqrt{KaV}$

(ii) Concentration [H+] of ion = $C\alpha = \sqrt{CKa}=\sqrt{\frac{Ka}{V}}$

iii) $pH=-\log\sqrt{CKa}=\frac{1}{2}(-\log Ka -\log C)$ ;

$pH=\frac{1}{2}(pKa -\log C)$ Increasing dilution, increases ionization and pH

### Mixture of strong acid and weak monoprotic acid

Let C1 and C2 be the concentrations of the strong and weak acids. If α is the degree of dissociation in the mixture, the hydrogen ion concentration = [H+] = C1+ C2*α.

Degree of dissociation of the weak acid will be less than the pure acid because of the higher [H+]

From the strong acid. This is referred as levelling effect. If the [H+] is less than 10-6 mole/l, hydrogen ion concentration from water also is to be added.

### Mixture of two Weak Monoprotic Acids

Say the two weak acid HA1 and HA2, have concentrations C1, C2 and degree of ionization α1 and α2. Initial concentration, moles/l

At equilibrium, moles /l C1(1- α1) C1α1+ C2α2 C1α1 C2(1- α2) C1α1+ C2α2 C2α2

So, $Ka=\frac{[H+][A-]}{[HA]}\;\;\;\;\;\;\;\;Ka1=\frac{c1\alpha 1(c1\alpha 1+c2\alpha 2)}{C1(1-\alpha)}\;\;\; Ka2=\frac{C2\alpha 2(1\alpha 1+C2\alpha 2)}{C2(1-\alpha2)}$

Since α is small, Ka1 = (C1α1+ C2α2) α1 Ka2 = (C1α1+ C2α22

$\alpha 1 = \frac{Ka1}{(C1\alpha 1+C2\alpha 2)}\;\;\;\;\alpha2=\frac{Ka2}{(C1\alpha1+C2\alpha2)}$ [H+] = C1α1+ C2α2 = $=\sqrt{C1 Ka1+ C2 Ka2}$