Question

Part A:

Given the two reactions

H2S⇌HS−+H+,   K1 = 9.51×10⁻⁸, and

HS−⇌S2−+H+,   K2 = 1.33×10⁻¹⁹,

what is the equilibrium constant Kfinal for the following reaction?

S2−+2H+⇌H2S

Part B:

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3 = 1.83×10⁻¹⁰, and

AgCl⇌Ag++Cl−,   K4 = 1.20×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

Answer 1

A)

For

H2S⇌HS−+H+

Ka1 = [HS-][H+]/[H2S]

and

HS−⇌S2−+H+

Ka2 = [S2-][H+]/[HS-]

then; we need

S2-+2H+ <-> H2S

the equilibriumn expression

Kt = [H2S]/ [H+]^2[S2-]

We need to move 1 and 2 in order to get Kt

add (1) and (2)

H2S⇌HS−+H+ Ka1

HS−⇌S2−+H+ Ka2

Add both

H2S + HS−⇌HS−+H+ + S2−+H+ Kt = (Ka1*Ka2)

cancel common terms

H2S ⇌ 2H+ + S2− Kt = K1*K2

but we need the reverse reaction so

invert it

K = 1/(K1*K2)

2H+ + S2−  ⇌ H2S

NOTE: there was a math error; it is 1.33*10^-19 and typing error stated 1.33*10^-9

Kt = (1/Ka1)*1/Ka2 = 1/(9.51*10^-8) * 1/(1.33*10^-19) = 7.906*10^25

B)=

PbCl2⇌Pb2++2Cl− K3 = 1.83*10^-10

AgCl⇌Ag++Cl− K4 = 1.2*10^-4

invert (2) and multiply by 2

that is

1/(K4^2)

Then

Knew = K3*1/(K4^2) = (1.83*10^-10)/(1.2*10^-4)^2

Knew = 0.012708