(i) pH = 5.5
pH = -log[H⁺]
= 5.5 -log [H⁺]
or
-5.5 log [H⁺]
= 6.5000
= Antilog of [-6 + 0.5]
= 3.162 x 10^-6
[H⁺] = 3.162 x 10^-6
(ii) pH = 8.6
= 3.162 x 10^-6
8.6 = -log [H⁺]
log [H⁺] - 8.6
[H⁺] = antilog of [-9 + 0.400]
= 2.512 x 10^-9
[H⁺] = 2.512 x 10^-9
(iii) pH = 3.2
3.2 = -log [H⁺]
log [H⁺] = -32
[H⁺] = antilog of [-4 + 0.8]
= 6.31 x 10^-4
[H⁺] = 6.31 x 10^-4

The degree of ionisation and the dissociation constant of the weak acid are related by the equation.
\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } \cong { \alpha }^{ 2 }C\)
α = 5% (or) 0.05
C = 0.02M
K_a = (0.05)2 x 0.02 = 0.00005
K_a = 5 x 10^-5

\(N{ H }_{ 4 }\rightleftharpoons { N{ H }_{ 4 } }^{ + }+{ OH }^{ - }\)
If α is the degree of dissociation, and C is the concentration in moles/lit. then
\({ K }_{ b }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =\frac { 0.04\times 0.042\times 0.01 }{ -0.042 } \)
K_b = 1.84 x 10^-5.

\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =\frac { .02\times .02\times 0.1 }{ 1-0.02 } \)
\({ K }_{ a }=\frac { 0.4\times { 10 }^{ -4 } }{ 0.98 } =4.08\times { 10 }^{ -5 }\)

(HCI strong acid fully ionised)
\(pH=\log\frac { 1 }{ { H }^{ + } } =\log\frac { 1 }{0 .01 } \)
\(=\log\frac { 100 }{ 2 } =1.6990\)

pH = -log [H⁺]
∴ [H⁺] = antilog [-pH]
= antilog [-7.41]
∴ [H⁺] = 3.9 x 10^-8 M.

\(pH={ pK }_{ a }+{ \log }\frac { [salt] }{ [acid] } \)
The concentration of the salt and the acid should be in moles/lit.
Number of moles of acetic acid
= \(\frac{weight \ of \ acetic \ acid}{molecular \ weight \ of \ acetic \ acid}=\frac{7.5}{60}\)
Number of moles of sodium acetate
= \(\frac{weight \ of \ sodium \ acetate}{molecular \ weight \ of \ sodium acetate}=\frac{10.25}{82}\)
pK_a = -log K_a
= -Iog 1.8 x 10^-5
\(=log\frac { 1 }{ 1.8\times { 10 }^{ -5 } } \)
= log 1 - log 1.8 - log 10^-5
= 5 - 0.2553 = 4.7447
\(\therefore pH=4.7447+\log\frac { 0.12 }{ 0.08 } \)
pH = 4.7447.

HCI ⟶ H⁺ + Cl^-. HCI is a strong acid.
[H⁺] from HCI is very much greater than [H⁺] from water which is 1 x 10^-7 M.
∴ [H⁺] = [HCI] = 0.001 M
∴ pH = -log (0.001) = 3.0
∴ That is acidic solution.

Formic acid is a weak acid. Let it be represented as HA.
\(\underset { C(1-\alpha ) }{ HA } \rightleftharpoons \underset { C\alpha }{ { H }^{ + }+ } \underset { C\alpha }{ { A }^{ - } } \)
According to Ostwald's dilution law;
\({ K }_{ a }={ C\alpha }^{ 2 }\)
\(\therefore \alpha =\sqrt { \frac { { K }_{ a } }{ C } } \)
K_a = 1.8 x 10^-4; C = 0.2 M = 2 x 10^-1 M
\(\therefore \alpha =\sqrt { \frac { 1.8\times { 10 }^{ -4 } }{ 2\times { 10 }^{ -1 } } } =\sqrt { 9\times { 10 }^{ -4 } } =3\times { 10 }^{ -2 }\)
Percentage of ionisation = 100 α
= 10² x 3 x 10^-2 = 3

Ba(OH₂) ⟶ Ba²+ 2OH^-
∴ [OH^-] = 2 [Ba(OH)²]
= 2 x 0.02 = 0.04 M
∴ pOH = -log [OH^-]
= 1.398 = 1.40
∴ pH = 14 - 1.4 = 12.6

pH = - log [H₃O⁺]
(i) pH - log [10⁴]
pH = log 1 - log 10⁴
pH =  - 4
(ii) pH = - log [10^-7]
pH = log 1 - log 10^-7.
pH = 7
(iii) pH = - log [H₃O⁺]
pH = - log [6.8 x 10^-3]
pH = log 1 - log 6.8 - log 10^-3
= 3 - 0.8325
pH = 2.17 (or) 2.2
(iv) pH = - log [3.2 x 10^-5]
pH = log 1 - log 3.2 - log 10^-5
= 5 - 0.5051
pH = 4.49 (or) 4.5
(v) pH = - log [0.035]
pH = log 1 - log 0.035
= 2 - 0.5441
pH = 1.46 (or) 1.5
(vi) pH = - log [0.25]
pH = log 1 - log 0.25
= 1 - 0.3979
pH = 0.602 (or) 0.60
(vii) pH = -log [H₃O⁺]
pH = log 1 - log 5.4 - log 10^-9
= 7 - 0.7324
pH = 8.267 (or) 8.3
(viii) pH = - log [7.1 x 10^-7]
pH = log 1 - log 7.1 - log 10^-7
= 7 - 0.8513
pH = 6.2.

pH + pOH = 14.0
∴ pOH = 14.0 - pH
= 14.0 - 5.5 = 8.50
pOH = 8.5 ∴ antilog [-pOH]|
∴  [OH^-] = antilog [-8.5] = 3.2 x 10^-9 M.

Degree if dissociation \(\alpha =\sqrt { \frac { { K }_{ B } }{ C } , } C\alpha =\sqrt { { K }_{ a }.C } \)
∴ \(pH=\log\frac { 1 }{ \sqrt { { K }_{ a }.C } } =\log\frac { 1 }{ \sqrt { 1.8\times { 10 }^{ -5 }\times .1 } } \)
= \(\log\frac { 1 }{ \sqrt { 1.8\times { 10 }^{ -3 } } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.8 } } \)
\(=3-\frac { 1 }{ 2 } \log1.8\)
\(=3-\frac { 1 }{ 2 } \times 0.2553=3-0.1276=2.8724\)
Alternating,
\(\left[ { H }^{ + } \right] =\sqrt { { K }_{ a }\times C } \)
\(=\sqrt { 1.8\times { 10 }^{ -5 }\times 0.1 } =\sqrt { 1.8\times { 10 }^{ -6 } } \)
= 1.341 x 10^-3
∴ -log [H⁺] = -log (1.341 x 10^-3)
= 3 - 0.1277 = 2.8723.

The pH of a buffer solution is given by the Henderson's equation.
\(pH={ pK }_{ a }+{ \log }\frac { [salt] }{ [acid] } \)
pK_a = -log K_a = -log [1.8 x 10^-5]
\(=\log\frac { 1 }{ 1.8\times { 10 }^{ -5 } } \)
= log 1 - log 1.8 - log 10^-5
= 5 - 0.2553
= 4.7447
\(pH=4.7447+\log\frac { 0.12 }{ 0.08 } \)
= 4.7447 + log 1.5
= 4.7447 + 0.1761
pH = 4.921.

The definition of pH is
We are given [H⁺] = 3.3 x 10^-2
Substituting into the definition of pH, we get
pH = -log (3.3 x 10^-2)
= - (- 1.48) = 1.48
Since the pH is less than 7.00, the solution is acidic.