Chem 1412 1/Calculate the solubility (in g/L) of Ag 2 CO 3 . (K sp =8.1X10 -12 )

Question

Chem 1412

1/Calculate the solubility (in g/L) of Ag2CO3. (Ksp=8.1X10-12) (molecular mass: Ag=107.9, C = 12, O = 16)

2/What mole ratio of KC2H3O2 to HC2H3O2 would buffer a solution at pH 4.54? (Ka for HC2H3O2 is 1.8X10-5)

3/For a particular reaction, DeltaH = -32kJ and DeltaS = -98J/K. Assume that deltaH and deltaS do not vary with temperature. (a) at what temperature will the reaction have deltaG=0? (b) if temperature is increased from that in part (a), will the reaction be spontaneous or non-spntaneous?

4/The equilibrium constant (Kp) for the reaction

H2(g) + CO2(g) => H2O(g) + CO(g)

is 4.40 at 2000K. (a) Calculate deltaGo for the reaction. (b) calculate deltaG for the reaction when the partial pressures are PH2 = 0.25 atm, PCO2 = 0.78 atm, PH2O=0.66 atm and PCO=1.20 atm and predict the direction of the reaction

5/ What is the solubility of AgCl in (a) H2O, (b) 0.0010 M AgNO3 solution? Ksp for AgCl is 1.8x10-10

6/ Calculate the solubility of Mg(OH)2 in a solution with a pH of 12.8 (Ksp of Mg(OH)2 is 1.2x10-11)

7/ For a reaction to be spontaneous at any temperature, which one of the following is TRUE

(a) deltaH is possitive and deltaS is positive

(b) deltaH is possitive and deltaS is negative

(c) deltaH is negative and deltaS is negative

(d) deltaH is negative and deltaS is positive

8/ Which one of the following solutions can act as a buffer?

(a) NaI/HI

(b) NaOH/NaCl

(c) NaCN/HCN

(d) NaI/CH3COONa

9/ A Student is asked to prepare a buffer solution at pH = 3.6, which one of the following weak acids should be used?

(a) NH4OH (Kb=1.8x10-5)

(b) HA (Ka=2.7 x10-4)

(c) HB (Ka=1.5x10-9)

(d) HC (Ka=5.3x10-13)

10/ Which one of the following statement is FALSE?

In the equilibrium Mg(OH)2(s)=> Mg2+(aq) + 2OH-(aq)

(a) the reaction moves forward when adding in acid

(b) The reaction moves backward when adding in solid NaCl

(c) The reaction moves backward when adding in solid MgCl2

(d) The reaction moves forward when adding in water

11. Exactly 100mL of 0.10 M HNO2 are titrated with a 0.10 M NAOH solution. Calculate the pH for (a) the initial solution, (b) the point at which 80 mL of the base has been added, (c) the equivalence point, (d) the point at which 105 mL of the base has been added. Ka=4.5x10-4

Explanation / Answer

11)HNO2 <----> H+ + NO2-
Ka = 4.5 x 10^-4 = x^2 / 0.11 -x
x = [H+] = 0.00704 M
pH = 2.15

Moles HNO2 = 100 x 0.11 / 1000 = 0.011
Moles NaOH = 80 x 0.11 / 1000 = 0.0088
0.011 mole HNO2 + 0.0088 mole NaOH >> 0.0022 mole HNO2 + 0.0088 mole NO2-
Total volume = 200 mL = 0.200 L
[HNO2] = 0.0022 / 0.200 = 0.011 M
[NO2-] = 0.0088 / 0.200 = 0.044 M
pKa = - log Ka = 3.35
pH = pKa + log 0.044 / 0.011 = 3.95

Moles HNO2 = moles NaOH = 0.011
Total volume = 0.200 L
[NO2-] = 0.011 / 0.200 = 0.055 M
Kh = 1 x 10^-14 / 4.5 x 10^-4 = 2.22 x 10^-11
NO2- + H2O <---> HNO2 + OH-
2.22 x 10^-11 = x^2 / 0.055 -x
x = 1.11 x 10^-6 M
pOH = 5.96
pH = 8.04

Moles NaOH = 105 x 0.11 / 1000 =0.01155
Moles HNO2 = 100 x 0.11 / 1000 = 0.011
Moles OH- in excess =0.00055
Volume = 205 mL = 0.205 L
[OH-] = 0.00055 / 0.205 = 0.00268 M
pOH = 2.57
pH = 11.43

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