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## SS1: CHEMISTRY - 2ND TERM

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Lesson 9, Topic 1
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# pH and pOH

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pH is a measure of hydrogen ion concentration [H+] or hydroxonium ion concentration [H3O+] in a solution. It is a measure of the strength of the acid or base character of a substance.

pOH is a measure of hydroxyl ion concentration [OH]

### Ionic Product of Water (Kw)

H2O(l) $$\rightleftharpoons$$ H+(aq) + OH(aq)

2H2O(l) $$\rightleftharpoons$$ H3O+(aq) + OH(aq)

The concentration of the two ions in water H+ & OH is 10-7

[H+] is the concentration of hydrogen ion

[OH] is the concentration of hydroxyl ion.

The product is

[H+] [OH] = 10-7 X 10-7 = 10-14

Kw is the ionic product of water or dissociation constant of water at 25ºC.

When [H+] = [OH], the liquid is neutral

[H+] = [OH] = 10-7 mol/dm3

The ionic product, Kw = [H+] [OH] = 10-14 enables pOH to record alkalinity as well as acidity.

### Calculation of pH and pOH:

pH is defined as the negative Logarithm to base 10 of hydrogen ion concentration [H+] expressed in moles per dm3.

pH = -Log10[H+], pH = – Log10 [H3O+]

pOH is defined as the negative Logarithm to base 10 of hydroxyl ion concentration [OH] expressed in moles per dm3.

pOH = -Log10 [OH]

pH + pOH = pKw

7 + 7 = 14

pH + pOH = pKw = 14

The pH scale ranges from 1 to 14, in which case, 1 is the lower limit and 14 is the upper limit.

An acidic solution has a pH value less than 7. As pH decreases, the acidity of the solution increases. Neutral solutions have a pH of 7. An alkaline solution has a pH greater than 7 and the level of alkalinity increases as the pH increases.

pH or pOH can be calculated if the concentration (in mol dm3) containing H+, H3O+, or OH is known. There are two important cases in this respect.

(a) Aqueous solutions of strong acid and bases. Here, ionization is assumed to be total in the production of H+, H3O+, or OH.

Examples are;

(i) HCl(aq) + H2O(l) $$\rightarrow$$H3O+(aq) + Cl(aq)

(ii) HCl(aq) $$\rightarrow$$H+(aq) + Cl(aq)

(iii) NaOH(aq) $$\rightarrow$$ Na+(aq) + OH(aq)

(b) By considering the appropriate mole ratio between the acid or base and the furnished ions (H+, H3Oor OH). The concentration of H+, H3Oor OH can be deduced, and hence, the pH or pOH of the solution calculated.

Example I:

Find the hydrogen ion concentration [H+] or hydroxyl ion concentration [OH] of the following

(i) 0.1M of HCl

(ii) 0.001M of NaOH

(iii) 0.01 M of H2SO4

(iv) 0.004M of KOH

(i) 0.1M of HCl

HCl(aq) $$\rightleftharpoons$$ H+(aq) + Cl(aq)

[H+] = 10-1

(ii) 0.001M of NaOH

NaOH(aq) $$\rightleftharpoons$$ Na+(aq) + OH(aq)

[OH] = 10-3

(iii) 0.01M of H2SO4

H2SO4(aq) $$\rightleftharpoons$$2H+(aq) + SO42-(aq)

0.01 x 2 = 2 x 10-2

(iv) 0.004M of KOH

KOH(aq) $$\rightleftharpoons$$K+(aq) + OH(aq)

[OH] = 4 x 10-3

Example II

Calculate the pH of the following aqueous solution:

(i) 0.001 mol dm-3 HCl

(ii) 0.001 mol dm-3 H2SO4

(iii) 0.001 mol dm-3 NaOH

(iv) 0.001 mol dm-3 Ba(OH)2

(v) 0.001 mol dm-3 NaCl

Solution

(i) 0.001 mol dm-3 HCl

HCl $$\rightleftharpoons$$ H+(aq) + Cl(aq)

[H+] = 10-3

pH = -Log10 [H+]

pH = – Log1010-3

= -1 x -3log1010 = 3

Note that log1010 = 1

(ii) 0.001 mol dm-3 H2SO4

$$\scriptsize \underset {1\;mol \;dm^{-3}}{H_2SO_{4(aq)}} \; \rightleftharpoons \; \underset {2\;mol \;dm^{-3}}{2H^+_{(aq)}} \; + \; \underset {1\;mol \;dm^{-3}} {SO_{4(aq)}^{-2}}$$

Since H2SO4 is a dibasic acid

[H+] = 2 x 10-3

pH = -log10[H+]

pH = -log10(2 x 10-3)

= 3log10 – log 2

= 3 – 0.3010

= 2.699

(iii) 0.001 mol dm-3 NaOH

NaOH $$\rightleftharpoons$$ Na+ + OH

[OH] = 10-3

pOH = -log10[OH]

pOH = -log1010-3

= -1 x -3log1010 = 3

For aqueous solution,

pH + pOH = PKw

7 + 7 = 14

pOH = 3

pH = 14 – 3

pH = 11

(iv) 0.001 mol dm-3 Ba(OH)2

$$\scriptsize \underset {1\;mol \;dm^{-3}}{Ba(OH)_{2}} \; \rightleftharpoons \; \underset {1\;mol \;dm^{-3}}{Ba^{2+}_{(aq)}} \; + \; \underset {2\;mol \;dm^{-3}} {2OH^-_{(aq)}}$$

Since Ba(OH)2 is a dibasic Base,

[OH] = 2 x 10-3

pOH = -log10[OH]

pOH = -log10(2 x 10-3)

=3log1010 – log102

=3 – 0.3010

pOH = 2.699

For aqueous solution

pH + pOH = PKw

7 + 7 = 14

pOH = 2.699

pH = 14 – 2.699 = 11.301

pH = 11.301

(v) 0.001 mol dm-3 NaCl

$$\scriptsize \underset {1\;mol\;dm^{-3}\\ 10\;mol\;dm^{-3}}{ Nacl_{(aq)}} \; \rightarrow \; \underset{1\;mol\;dm^{-3} \\ 10\;mol\;dm^{-3} }{Na^+{(aq)}} \; + \; \underset{1\;mol\;dm^{-3}\\ 10\;mol\;dm^{-3}}{ Cl^-{(aq)}}$$

Since there is no excess H3O+, H+ or OH in solution, the solution is neutral

Hence, [H+] = [OH] = 10-7 mol dm-3

pH = -log10[H+] = -log1010-7 = 7

Example III

The hydrogen ion concentration [H+] in a solution is 5 x 10-3 moles dm-3. Determine the pH of the solution.

[H+] = -5 x 10-3 moles dm-3

pH = -log10[H+]

= -log10 (5 x 10-3)

= -(log10 5 + log1010-3)

= -[log10 5 + (-3log1010]

= -[log10 5 + (-3 x 1)]

= -[log10 5 – 3]

= -log10 5 + 3

= -0.6990 + 3

pH = 2.30

Example IV

What is the [H+], [OH], pH, and pOH of the solution corresponding to the 0.02M NaOH?

Solution

[OH] = 0.02 mol dm-3

[H+][OH] = 10-14

[H+] = $$\frac {10^{-14}} {OH^{-}} \\ \frac {10^{-14}}{2 \; \times \; 10^{-2}}$$

[H+] = 0.5 x 10-12 mol dm3

pH = -log10[H+]

=-log10 (0.5 x 10-12)

= -(log10 0.5 + log1010-12)

-[log10 0.5 + (-12 log1010)]

-[log10 0.5 + (-12 x 1)]

=-[log10 0.5 – 12]

=-log10 0.5 + 12

= -(-0.3010) + 12

= + 12.3010

pH = 12.3010

pH + pOH = pKw

7   +   7     = 14

pH = 12.3010

pOH = 14 – 12.3010

= 1.6990

Example V

What is the hydroxyl ion concentration [OH] in a solution with a pH of 5.5?

Solution

pH + pOH = pKw

pH = 5.5

pOH = 14 – 5.5

pOH = 8.5

pOH = -log10[OH]

8.5 = -log10[OH]

-8.5 = log10[OH]

Antilog – 8.5 = [OH]

3 x 10-3 mol dm-3 = [OH]

Example VI

Calculate

(i) Hydroxonium ion concentration [H3O+

(ii) Hydroxyl ion concentration [OH]

of a solution with pH 6.5

Solution

(i) pH = -log10 [H3O+]

6.5 = -log10 [H3O+]

–6.5 = log10[H3O+]

Antilog –6.5 = [H3O+]

3.16 x 10-7 mol dm-3 = [H3O+]

(ii) [H3O+] [OH] = KW = 10-14

[H3O+] [OH] = 10-14

[OH]  = $$\frac {10^{-14}} {H_3O^{+}} \\ \frac {10^{-14}}{3.16 \: \times \: 10^{-7}}$$

[OH] = 0.32 x 10-7 moldm-3

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