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

AgI	⇌	Ag+	+	I-
s		s		s

K_sp = s²

$$ \Rightarrow $$ 1.0 × 10^–16 = s²

$$ \Rightarrow $$ s = 1.0 × 10^–8 mol L^–1

$$ \therefore $$ [Ag⁺] = 1.0 × 10^–8 mol L^–1

Also, in 10^–4 N KI solution,

[I^–1] = (10^–4 + 1.0 × 10^–8) mol L^–1

$$ \Rightarrow $$ [I^–1] = (10^–4) mol L^–1

[As 1.0 × 10^–8 mol L^–1 << 1.0 × 10^–4 mol L^–1]

$$ \therefore $$ K_sp of AgI = [Ag⁺][I^–]

= (1.0 × 10^–8)(10^–4)

= 1.0 × 10^–12 mol L^–1

For any reaction to process in forward direction the reaction quotient (Q) must be less than equilibrium constant K_C.

Q < K_c