Ionic Equilibrium

In HCl solution [H⁺] = 10^-8 M

Also from water, [H⁺] = 10^-7 M

$ \therefore $ Total [H⁺] = 10^-8 M + 10^-7 M

= 1.1 ×10^–7 M

$ \simeq $ 1.0525 $ \times $ 10^$-$7 M

HNO₂ is a weak acid and NaNO₂ is its salt thus these two constitute to form an acidic buffer.

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

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

NH₄OH and NH₄Cl constitute to form a basic buffer.

pOH = pK_b + log ${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$

We know, pOH+ pH = 14 or pOH = 14 – pH

$ \therefore $ 14 - pH - $\log {{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$ = pK_b

$ \Rightarrow $ 14 - 9.25 - $\log {{0.1} \over {0.1}}$ = pK_b

$ \Rightarrow $ 14 – 9.25 – 0 = pK_b

$ \Rightarrow $ pK_b = 4.75

MX2	⇌	Ag2+	+	2X-
s		s		2s

K_sp = [M²⁺] [X^–]² = (S)(2S)² = 4S³

$ \Rightarrow $ K_sp = 4(0.5 × 10^–4)³ = 5 × 10^–13

Na₂CO₃ is a salt of weak acid H₂CO₃ and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.

M₂S ⇌ 2M⁺ + S^2–

K_sp = [M⁺]²[S^2–] = (2s)²(s) = 4s ³

K_sp = 4(3.5 × 10^–6)³ = 1.7 × 10^–16

Strong base has higher tendency to accept the proton. Increasing order of base and hence the order of accepting tendency of proton is

R $-$ NH₂ > NH₃ > HS^$-$ > I^$-$

CH₃COOH ⇌ CH₃COO^– + H⁺

K_c = ${{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}}$

$ \Rightarrow $ [CH₃COOH] = ${{{3.4 \times {{10}^{ - 4}} \times 3.4 \times {{10}^{ - 4}}} \over {1.7 \times {{10}^{ - 5}}}}}$

= 6.8 × 10^–3

In polyprotic acids the loss of second proton occurs much less readily than the first. Usually the K_a values for successive loss of protons from these acids differ by at least a factor of 10^–3.

$ \therefore $ K_a1 > K_a2

NH₃ after losing a proton (H⁺) gives NH₂^–. So, NH₃ is the conjugate acid of NH₂^–.

Note : Conjugate acid-base pair differ only by a proton.

CH₃COOH is weak acid while NaOH is strong base, so one equivalent of NaOH can not be neutralized with one equivalent of CH₃COOH. Hence the solution of one equivalent of each does not have pH value as 7. Its pH will be towards basic side as NaOH is a strong base hence conc. of OH^– will be more than the conc. of H⁺.