6. Calculate the molar solubility of Mg(OII)_2 in a solution with a pH of 12.8 (K_op of Mg(OH)^2 is 1.2 * 10^-11
7. For a reaction to be spontaneous at any temperature, which one of the following is true:
a) [delta] H is positive and [delta] S is positive
b) [delta] H is positive and [delta] S is negative
c) [delta] H is negative and [delta] S is negative
8. Which one of the following solutions can act as a buffer?
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) NH_4CI (K_b = 1.8 * 10^-5 for NH_4OH
b) HA (K_a = 2.7 * 10 ^-4)
c) HB (K_a = 1.5 * 10^-9)
d) HC (K_a = 5.3 * 10^-13)
10. Which one of the following statements is false?
In the equilibrium Mg(OH)_ (s) ---Mg^2+ (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 MgCl^2
d) The reaction moves forward when adding in water
11. Acetic acid, CH_3COOH is a weak acid (K_a - 1.8 * 10 ^-5)
a) Solution A is 100ml of 0.10M acetic acid
b) Solution B is prepared by adding 0.0050 mol solid NaOH to Solution A
c) Solution C is prepared by adding 0.010 mol solid NaOH to Solution A
d) Solution D is made by adding 0.0150 mil solid NaOH into Solution A
e) Calculate the pH for each of the solutions, A,B,C, and D
SOLUTION This solution is FREE courtesy of BrainMass!
6. Mg(OH)2 -> Mg2+ + 2OH-
As pH = 12.8, pOH = 1.2 which means [OH-] = 10-1.2 = 0.063 M
If 's' is the molar solubility,
Ksp = (s)(2s + 0.063)2 = 1.2 x 10-11
As molar solubility will be really small compared to 0.063, hence 2s + 0.063 = 0.063 approximately
Hence, s(0.063)2 = 1.2 x 10-11
Solving for molar solubility, s = 3 x 10-9 M
7. [delta] G = [delta] H - T[delta] S
For spontaneity, [delta] G must be negative. This is possible at all T if [delta] H is negative and [delta] S is positive. I don't see the option given, so none of the options are true.
8. A buffer contains a weak acid and its salt or a weak base and its salt. Of all the options, only NaCN/HCN can be a buffer as it has a weak acid HCN and its salt NaCN.
9. You choose the weak acid whose pKa is closest to the desired pH. In this case, HA has a pKa = 3.57 and hence is the best choice.
10. The only statement that is false is 10b as the added NaCl will dissolve in water but will not change the concentration of either reactants or products.
11. a. Moles of acetic acid = 0.1L x 0.1M = 0.01 mol
If 'x' is the [H+], then for the equilibrium CH3COOH -> CH3COO- + H+
Ka = x2 / (0.1M - x) = 1.8 x 10-5
Solving for x = 0.00133M. Hence pH = -log[H+] = 2.876
b. Adding 0.0050 mol of NaOH to 0.01 mol CH3COOH will form 0.0050 mol of CH3COONa and leave 0.0050 mol CH3COOH unreacted. When the mol of weak acid = the mol of its salt, then pH = pKa. Hence pH = 4.74
c. Adding 0.01 mol NaOH will result in 0.01 mol of CH3COONa and no CH3COOH left. Now the equilibrium will be ,
CH3COO- + H2O -> CH3COOH + OH-
If 'x' is the [OH-], then
Kb = x2 / (0.1M - x) = 5.55 x 10-10 using the relation Ka x Kb = 1x 10-14
Solving for x, gives x = 7.454 x 10-6. Hence pOH = 5.128 and thus pH = 14-pOH = 8.872
d. Adding 0.015 mol NaOH leaves 0.005 mol NaOH unreacted which will control the pH.
[OH-] = 0.005 / 0.1 = 0.05M
pH = 14 - pOH = 12.7