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). H2SO4(aq)→ 2H+(aq)+SO42-(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.
OBJECTIVE QUESTIONS
1.
No comments:
Post a Comment