\(K_h= { \frac { { K }_{ w } }{ K_b } } = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\)
= 0.55 x 10^-9 = 5.5 x 10^-10

HPO₄²⁻ can have the ability to accept a proton to form H₂PO₄^-
It can also have the ability to donate a proton to form PO₄^-3

x ml of 0.1 M NaOH + x mL of 0.01 M HCI
No. of moles of NaOH = 0.1 x X x 10^-3
= 0.1 X x 10^-3
No. of moles of HCI = 0.01 x X x 10^-3
= 0.01 X x 10^-3
No. of moles of NaOH after mixing
= 0.1 X x 10^-3 - 0.01 X x 10^-3
= 0.09 X x 10^-3
Concentration of NaOH\(= (\frac{0.09x \times 10^{-3}}{2x \times 10^{-3}}) = 0 .045\)
[OH^-] = 0.045
pOH =-log (4.5 x 10^-2)
= 2 -log 4.5
= 2 - 0.65 = 1.35
pH = 14 - 1.35 = 12.65

AgCl_(s) ⇌ Ag⁺_(aq) + Cl^-_(aq)
NaCl  → Na⁺ +Cl^-
0.1M      0.1M   0.1M
K_sp = 1.6 × 10^-10
K_sp = [Ag⁺][ Cl^-]
K_sp = (s) (s + 0.1)
0.1 >>> s
∴ s + 0.1 = 0.1
∴ s = 1.6 × 10^-10/ 0.1 = 1.6 × 10^-9

C₅H₅N + H-OH  ⇌  C₅H₅ ⁺NH + OH^-
\(\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =K_b\)
\(\alpha\)²C \(\approx\) K_b
\(\alpha = \sqrt {\frac{K_b} C} = \sqrt {\frac{1.7 \times 10^{-9}} {0.1}}\)
\(= \sqrt{1.7} \times 10^{-4}\)
Percentage of dissociation =\(= \sqrt{1.7} \times 10^{-4}\) x 100
= 1.3 x 10^-2 = 0.013%

BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CF₄ → neutral →  neither Lewis acid nor base
SiF₄ → neutral →  neither Lewis acid nor base

H₂O + H₂O ⇌  H₃O⁺ + OH^-
acid 1  base 1  acid 2   base 2
HF + H₂O ⇌  H₃O⁺ + F^-
acid 1  base 1  acid 2   base 2
∴ Conjugate bases are OH^- and F^- respectively.

\(\mathrm{Ag}_{2} \mathrm{C_2O_4} \rightleftharpoons 2 \mathrm{Ag}_{}^{+}+\mathrm{C_2O}_{4{}}^{2-}\)
\(\left[\mathrm{Ag}^{+}\right]=2 .24 \) ×10^-4mol L^-1
\(\mathrm{C_2O}_{4{}}^{2-} = \frac {2.24 \times 10 ^{-4}}{2}\) mol L^-1
= 1.12 ×10^-4mol L^-1
K_sp = [Ag]² [C₂O₄^2-]
= (2.24 ×10^-4mol L^-1) (1.12 ×10^-4mol L^-1)
= 5.619 × 10^-12mol³L^-3