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SOLUTIONS

1.

System	Add	How would vary ?
CaCO₃(s) + H₂O closed	CaCO₃ (s)	[Ca⁺²] = C_T=
H₂O	ZnO(s)	pH -
CO₂ (open)	NaCl	Alk = C_T =
H₂O	SO₂	pH - (acid)
CaCO₃(s)	FeCl₃	Ca⁺² + acid salt
Fe⁺³/Fe⁺² at E=1V	Decrease E	Fe⁺³/Fe⁺² - more reducing
Fe⁺³/Fe⁺² in the presence of H₂S	O₂	E + addition of Fe⁺³/Fe⁺² + oxidant

2.

7; 0.7; 3.7

3.

Reduction: Zn⁺² + 2 e^- Zn (s) E^o = -0.76 V

Oxidaton: Pb(s) Pb⁺² + 2 e^- E^o = 0.13 V

Overall: Zn⁺² + Pb(s) Zn(s) + Pb⁺² DE^o = -0.66V

Since the calculated standard cell potential is negative, lead will not reduce Zn⁺² ions.

4.

The following reaction control the solubility of metal carbonates:

MeCO₃ (s) Me⁺² + CO₃^-2 K_so

CO₂ (g) + H₂O H₂CO₃^* K_H

H₂CO₃^* H⁺ + HCO₃^- K_a1

HCO₃^- H⁺ + CO₃^-2 K_a2

K_so = [Me⁺²] [ CO₃^-2]

[CO₃^-2 ] = K_a2 K_a1 [H₂CO₃^*]/[H⁺]² = K_a2 K_a1 K_H P_CO2 /[H⁺]²

[Me⁺²] = K_so [H⁺]² / K_a2 K_a1 K_H P_CO2

If the pH is kept constant, as we increase P_CO2, the metal concentration decreases. In general, as we increase P_co2, the pH decreases and Me⁺² increases.

5.

NaOCl + H₂O ----------> Na⁺ + OCl^-

OCl^- + H₂O HOCl + OH^-

Species: Na⁺, OCl^- , HOCl, OH^-, H⁺

MB : [OCl^- ] + [HOCl] = C_T = 10^-3

PBE: [HOCl] + [H⁺] = [OH^- ]

Assumption: as we add a base [OH^-] >>> [H⁺]

Thus in PBE : [HOCl] = [OH^- ]

In MB: [OCl^- ] = 10^-3 - [HOCl] = 10^-3 - [OH^- ]

At pH = 9.2, [HOCl] = [OH^- ]= 1.97x10^-5 M ;;;;;;;;;;

[OCl^-] = 1x10^-3 - 1.97x10^-5 ~ 1x10^-3 M

We need to add acid to reach pH = 7.5

At pH = 7.5, we can not follow the PBE as we do not know the type of acid added but:

At pK_a = pH

HOCl]= [OCl^-]
[HOCl]= [OCl^-]=C_T/2 = 5.0 x 10^-4 M
C_T = [HOCl]= [OCl^-]

We can solve the problem graphically

6.

The two half-cell potentials are:

We write the oxidation and the reduction reaction with the same number of electrons in each side:

And we compute the standard cell potential as:

(b)

7.

H⁺ ; OH^-; M⁺² ; CO₃^-2 ; HCO^-₃; H₂CO₃

Is an open system with a fixed P_CO2 and a metal carbonate present:

Electroneutralty equation:

[H⁺] + 2[M⁺²] = [OH^-] + [HCO₃^-]+ 2[CO₃^-2]

Assumptions: As this is common in natural systems, pH is close to 7 and then:

[H⁺] << [M⁺²]

[HCO₃^-]>>[CO₃^-2]

And from the ENE:

2[M⁺²] = [HCO₃^-]

2(10^12.6)[H⁺]² = 10^-11.3/[H⁺]

Solve for [H⁺] = 10^-8.07 pH = 8.07

look at the graph to see that assumptions are correct

Alk = [OH^-] + [HCO₃^-] + 2[CO^-2₃] - [H⁺] ~ = [HCO₃^-] = 10^-11.3/10^-8.07 = 10^-3.23

CO₂ (g) + MCO₃ (s) + H₂O + H⁺ M⁺² + 2H₂CO₃^*

H⁺ is consumed, alkalinity decreases.