H₃PO₄ + H - OH ⇌ H₃O⁺ + H₂PO₄^-
acid 1 base 1         acid 2 base 2
\(\therefore\) H₂PO₄^- the conjugate base of H₃PO₄ 

PbI_2(s) ⇌  Pb²⁺_(aq)  + 2I^-_(aq) 
K_sp = (s) (2s)²
3.2 × 10^-8 = 4s³
s = (3.2 × 10^-8/4)^1/3
= ( 8 x  10^-9)^1/3
= 2 x 10^-3 M

pH = -log₁₀ [H⁺]
∴  [H⁺] = 10^-pH
Let the volume be x ml.
V₁M₁ +V₂M₂ +V₃M₃ = VM
∴   x ml of 10^-1 M+ x ml of 10^-2 M + x ml of 10^-3 M
= 3 x ml of [H⁺]
\(∴  [H^+] = \frac{ x[0.1 + 0.01 + 0.001]}{3x}\)
\(= \frac{ [0.1 + 0.01 + 0.001]}{3}\)
\(= \frac{ [0.111]}{3}\)
= 0.037
= 3.7 x 10^-2

HCOONa + HOH  ⇌ NaOH + H-COOH
                               strong base weak acid
Basic in nature.
C₆H₅NH₃Cl + HOH  ⇌ H₃O⁺ + C₆H₅ - NH₂ + Cl^-
                                      acidic
KCN + H - OH ⇌ KOH + HCN
Base                  strong base weak acid
basic, acidic, basic is correct.

BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CO → having lone pair of electron →  Lewis base
F → unshared pair of electron →  Lewis base

Basic buffer is the solution which has weak base and its salt.
NH₄OH + HCI → NH₄CI +H₂O + NH₄OH
200 ml   100 ml    salt       100 ml  weak base

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10 ^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)²  = 0.5 x 10^-15

75 mL\(\frac{M}{5}\)HCl + 25mL\(\frac{M}{5}\)NaOH
No. of moles of HCI = 0.2 x 75 x 10^-3 = 15 x 10^-3
No. of moles of NaOH = 0.2 x 25 x 10^-3
= 5 x 10^-3
No. of moles of HCl after mixing
= 15 x 10^-3 - 5 x 10^-3
= 10 x 10^-3
\(\therefore\) Concentration of HCI = No. of moles of HCI/Vol in litre
= 10 x 10^-3/100 x 10^-3 = 0.1 M
For (iii) solution. pH of 0.1 M HCI = -log₁₀ (0.1) = 1