# pH of Mixtures of Acids & Bases

Let us take x millimoles of acid (HA) and y millimoles of base (BOH). Note that acid is monobasic and base is monoacidic.

### (a) Strong acid & Strong base

#  If x = y; then complete neutralization takes place and we get x(=y) millimoles of salt (BA) of strong acid and strong base which means no hydrolysis takes place and pH of solution = 7.
# If x > y; then there is an excess of strong acid and resulting solution is acidic with millimoles of acid left in excess = x – y.
Now if V c.c. be the volume of mixture, then:

$\large M = \frac{x-y}{V}$

Now calculate pH using the equation pH = – log [H+]

#  If x < y; then there excess of strong base and resulting solution is basic with millimoles of base = y – x.

### (b) Strong base & weak acid

# If x = y; first of all neutralization takes place to give x(=y) millimoles of salt (BA). The salt will now undergo hydrolysis to give an alkaline solution. Calculate pH by using standard result:

$\large p_H = 7 + \frac{1}{2}(p_{K_a} + logC)$ ; C is concentration expressed in M (mol/lt)

# If x > y; there is excess of weak acid whose millimoles = x – y and y millimoles of salt is formed. This will give an acidic buffer solution. Calculate pH of buffer solution using Henderson’s Equation.

$\large p_H = p_{K_a} + log\frac{[salt]}{[acid]} = p_{K_a} + log\frac{y}{x-y}$

# If x < y; the solution in this case contains excess of strong base whose millimoles are y – x.

$\large M = \frac{y-x}{V}$ ; Calculate pH.

### (c) Strong acid & Weak base

# If x = y; first of all complete neutralisation takes place to produce x(=y) millimoles of salt (BA). The salt (BA) is of strong acid and weak base, hence hydrolysis takes place to give an acidic solution. Calculate its pH by using standard result.

$\large p_H = 7 – \frac{1}{2}(p_{K_b} + logC)$ ; C: is concentration of salt.

# If x > y; then solution contain excess of strong acid whose millimoles = x – y.

$\large M = \frac{x-y}{V}$ ; Calculate pH.

# If x < y; then there is an excess of weak base whose millimoles are y – x and millimoles of salt (BA) are x. This will give a basic buffer solution. Calculate the pH by using Henderson’s Equation.

$\large p_H = 14 – p_{K_b} – log\frac{[salt]}{[base]} = 14 – p_{K_b} – log\frac{x}{y-x}$

### (d) Weak acid & weak base

# If x = y; neutalisation takes place completely with the formation of x (=y) millimloles of salt (BA) of weak acid and weak base. So hydrolysis takes place. Calculate the pH by using:

$\large p_H = 7 + \frac{1}{2}(p_{K_a} – p_{K_b})$

# If x > y; then excess of weak acid (x – y) will remain with y millimoles of salt. This will give an salt buffer with acidic pH.

# If x < y; then excess of weak acid (y-x) and x millimoles of salt. This will give a salt buffer with basic pH.

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