pH at a glance

 

DEFINITION OF pH

pH is defined as the negative logarithm to the base 10 of the hydrogen ion [H⁺] concentration.

It is also defined as the degree of acidity or alkalinity of a solution

i.e. pH = -log₁₀ [H⁺].

Thus: If [H⁺] = 0.00001 or 10^-5.

log [H⁺] = log10^-5 = -5

pH= -log [H⁺] = - (-5) = 5.

If [H⁺] =10^-x

Therefore, pH= -log10^-x = - (-x) = x

If [H⁺] = 10^-2, pH = 2

 

DEFINITION OF pOH

pOH is defined as the negative logarithms of the hydroxide ion [OH^-] concentration to the base of 10.

i.e. pOH= -log [OH^-].

A solution with pH 7 is neutral.

A solution with pH less than 7 i.e. pH 6,5,4, 3, 2, 1 or 0 indicates increasing acidity as the numbers decreases.

A solution with pH greater than 7, i.e. pH 8,9,10, 11, 12, 13, or 14 indicate increasing alkalinity as the numbers increase.

 1< 2 <3< 4< 5< 6        7          8< 9<10< 11<12<13<14

Increasing acidity    Neutral     Increasing alkalinity

 

A solution with pH 1 is very acidic [with high concentration of H⁺]. A solution with pH 13 is very alkaline [with low concentration of H+, but high concentration of OH-].

Note that: If pH is 1, it has concentration of H+ 10 times greater than pH 2 and 100 times greater than pH 3 e.t.c.

pH 1 > pH 2 > pH 3.

Concentration of H⁺    10^-1      10^-2      10^-3.

                                      0.1       0.01     0.001.

 

Relationship between pH and pOH.

                 H₂O →   H⁺    +   OH^-

From conductivity measurement [H⁺] =10^-7moldm^-3, [OH^-]=10^-7moldm^-3.

[H⁺] [OH^-] = Kw=10^-7 x 10^-7=10^-14mol²dm^-6.

Taking logarithm of both sides

log ([H⁺] [OH^-]) = logKw

log [H⁺] + log[OH^-] =logKw

Subtracting both sides

-(log[H⁺]  +  [OH^-]) = -logKw

-log [H+] – log[OH^-] = -logKw

-log [H⁺] + (-log [OH^-]) = -logKw

pH + pOH = PKw

pKw = -log10^-14 = -(-14) = 14

Therefore, pH + pOH = 14.

 

Worked examples

1. Find the hydrogen and hydroxide ion concentrations in

(a) 0.01moldm^-3 tetraoxosulphate (vi) acid solution.

(b) 0.001moldm^-3 potassium hydroxide solution.

 

Solution

(a).  H₂SO_4(aq)→ 2H⁺_(aq)+SO₄^2-_(aq)

 

From the equation, 1 moldm^-3 H₂SO₄ ionizes to give 2moldm^-3 H⁺

Therefore, 0.01moldm^-3 H₂SO₄ would ionize to give (2x0.01) moldm^-3 H⁺

[H⁺] = 2x10^-2moldm^-3

[H⁺] [OH^-] = 10^-14

(2x10^-2) [OH^-] = 10^-14

[OH^-] =    10^-14

                2 x 10^-2

[OH-] = 0.5x (10^{-14- -2})

[OH-] =0.5 x10^-14+2

[OH-] =0.5x10^-12moldm^-3.

 

(b).     KOH_(aq)→ K⁺_(aq)+  OH^-_(aq)

From the equation,

1moldm^-3 of KOH ionizes to give 1moldm^-3 of OH-

10^-3moldm^-3 of KOH would ionize to give 10^-3moldm^-3 of OH

[OH^-]=10^-3moldm^-3.

[H⁺] [OH^-]=10^-14

[H⁺] (10^-3) = 10^-14.

[H⁺] = 10^-14

            10^-3

[H+] = 10^-14+3

[H+] = 10^-11moldm^-3

 

2.  A glass cup of orange juice is found to have a POH of 11.40. Calculate the concentration of the hydrogen ions in the juice.

Solution

pH + pOH = 14.

pH = 14 – 11.4.

pH = 2.6.

pH = -log [H⁺]

2.6 =-log [H⁺].

[H+] = Antilog (-2.6)

[H+] = 0.0025moldm^-3

[H+] = 2.5x10^-3moldm^-3.

 

Measuring pH of a solution.

We use the pH meter or a universal indicator to measure the pH of a solution. 

A universal indicator shows different colours at specific pH or hydrogen ion concentrations.