Answers to Questions. Page 345-346 #1 to 20. Exercise Answers Page 345.
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I am assuming that the author meant 1.0 moles of ammonia, and 0.30 moles of hydrogen. Otherwise there would only be one significant digit, and I never give chemistry problems with less than 2 significant digits.
1 mol in a 0.5 L flask =1.0/0.5 =2.0mol/L
0.3 mol in a 0.5 L flask =0.3/0.5 =0.6mol/L
2NH3 N2 + 3H2
a) The [N2] at equilibrium is 0.2 mol/l, the [NH3] is 1.6 mol/L
b) The equilibrium constant is :
Kc = [N2] [H2]3/ [NH3]2
or 0.20(0.60)3/ 1.62
so, Kc= 0.016875 or, rounded to 2 S.D: 1.7 x 10-2
SO2 + NO2 NO + SO3 Kc=4.8
1 11 1
b) At equilibrium:
Kc = [NO] [SO3]
4.8 = X2
4.8 (0.1296 - 0.72X + X2 ) = X2
0.622 – 3.46X+4.8X2=X2
3.8X2 -3.46X+0.622 = 0
X=0.247 mol/L in a 5L container
Total SO3 = 1.2 mol
10. (for this question, I am assuming the textbook intended to write 3.0 mol/L)
a) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
b) if [NH3]=[NO] then these two will cancel out in the equilibrium expression, so
Answer: The equilibrium constant is 0.26
2NO2(g) N2O4(g) Kc=1.15
a) The equilibrium expression is:
b) so x = 1.15 (0.05)2
x = 2.875 × 10-3mol/L or2.9 × 10-3mol/L
c) decreasing the pressure (by giving the reactants a larger volume) or increasing the concentration of N2O4 would shift the equilibrium towards the reactants.
a) HF(aq) + SO32-(aq) F-(aq) + HSO3-(aq)
b) CO32-(aq) + CH3COOH(aq) CH3COO-(aq) + HCO3-(aq)
c) H3PO4-(aq) + OCl-(aq) H2PO4(aq) + HOCl(aq)
d) HCO3(aq) + HSO4-(aq) SO42-(aq) + H2CO3(aq)
13. The answer is (a)
explanation: hydronium (H3O+) is a proton (H+) that has been hydrated (ie. combined with water: H+ +H2OH3O+)
14. The answer is (b)
explanation: Calculate pOH: -log[OH-] = -log(2.1) = -0.322
Calculate pH : 14-(-0.322) = 14.322
Calculate [H+] : 10-pH = 10-14.322 = 4.8 x 10-15
15. The answer is (b)
explanation: pOH = 14 – pH, so 14 – 3.46 = 10.54
16. The answer is (b)
explanation: calculate moles of NaOH 5g / 40.0 g/mol =0.125 mol
calculate concentration: 0.125mol / 4L = 0.3125 mol/L
calculate pOH:-log(.3125) = 1.505
calculate pH: 14 – 1.505= 12.495
17. Use the general formula:
HA + H2O H3O+ + A-
or simplified: HA H+ + A-
This is a weak acid, only five percent ionizes,
so 5% of a 0.10 mol/L
= 0.05 x 0.10 mol/L
=0.005 mol/L of H+ (or H3O+) ions and A- ions
The acid is 95% undissociated, so its concentration at equilibrium is 0.95 x 0.10 = 0.095 mol/L
Ka = [H+][A-] =(0.005)(0.005) = 2.6x10-4
The Ka of the weak acid is 2.6x10-4
Note: for very weak acids (less than 1% ionization) we don’t usually bother changing the denominator. Just leave the original concentration there, and your answer will still work out so close to correct that it will be the same after rounding Sig. Digits.
So, we just have to substitute the five sets of values to see which work out to 4.8x10-3
a) (1.0x10-4)2 ÷ 4.8x10-1 = 2.08x10-8
b) (4.8x10-4)2 ÷ 1.0x10-1 = 2.30x10-6
c) (2.2x10-2)2 ÷ 1.0x10-1 = 4.84x10-3
d) (1.0x10-1)2 ÷ 2.2x10-2 = 4.55x10-1
e) (1.1x10-2)2 ÷ 5.0x10-2 = 2.42x10-3
So the best answer is (c).
20. The best answer is (a)
explanation: (a) was the only case where water gave away its H+ (donated a proton). It gave the proton to the NH3 molecule, thus changing it into an NH4+ ion. According to the Bronsted-Lowry theory, any substance that donates a proton must be an acid.
In (b) it absorbed a hydrogen it got from the phosporic acid (accepted a proton), so it was acting like a base.
In (c) it decomposed into two covalent compounds,
In (d) the water did not change,
In (e) there was a chemical reaction, but there is no clear indication of any acid or base behavior.
21. Answer (b)
22. Answer: pH=11.08 or pH=11.1 (rounded to reasonable sig. Digits)
23. The soft drink is acidic because its pH is 2.3
24. Answer: c – b – a – e – d
25. Answer: (a) the pH of tears is 7.4
(b) the pH of gastric acid is 1.4
26. At equilibrium, the concentrations of [H2] and [I2] are both 2.5x10-4 mol/L
27. The system is not at equilibrium. Since no solid solute exists, the reverse reaction is not occurring.
28. The pH of codeine solution is 10.3
29. The calculation of the equilibrium constant uses stoichiometric coefficients. If the equation is not balanced, the coefficients will be incorrect.
35. Sulphur trioxide and hydrogen fluoride
The initial concentrations are found using C=n/V for SO3 and HF. They are [SO3]=0.617 mol/L and [HF]=1.936 mol/L
Using an ICE Table:
We could substitute the expressions in the bottom row into the equation, but that is about as far as we can go with this.
c. It is difficult to solve this because it contains terms that are to the sixth power, which cannot be solved by simple mathematical equations.