H₂S ⇌ H⁺ + HS^–

HCl ⇌ H⁺ + Cl^–

In presence of HCl this ionization of H₂S is suppressed due to the presence of extra H⁺ ions from HCl and produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H₂S which is sufficient to precipitate II^nd group radicals.

	BOH	⇌	B+	+	OH-
Initially	C		0		0
At equilibbrium	C - C$$\alpha $$		C$$\alpha $$		C$$\alpha $$

[OH^-] = C$$\alpha $$

$$ \Rightarrow $$ [OH^-] = $$\sqrt {{K_b}C} $$

$$ \Rightarrow $$ [OH^-] = $$\sqrt {1 \times {{10}^{ - 12}} \times {{10}^{ - 2}}} $$

= 1.0 $$ \times $$ 10^$$-$$7 mol L^$$-$$1

For an acid-base indicator

H_In ⇌ H⁺ + In^-

K_in = $${{\left[ {{H^ + }} \right]\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

$$ \Rightarrow $$ $$\left[ {{H^ + }} \right] = {{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

Take – log on both sides

$$ - \log \left[ {{H^ + }} \right] = - \log \left( {{{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}} \right)$$

$$ \Rightarrow $$ pH = –log K_In + $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

$$ \Rightarrow $$ pH = pK_In + $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

$$ \Rightarrow $$ $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$ = pH - pK_In

AX2	⇌	A2+	+	2X-
s		s		2s

K_sp = = [A²⁺] [X^– ]² = s × (2s)² = 4s³

$$ \Rightarrow $$ 3.2 $$ \times $$ 10^$$-$$11 = 4s³

$$ \Rightarrow $$ s³ = 8 × 10^–12

$$ \Rightarrow $$ s = 2 × 10^–4 mol L^–1