1. The following question refers to a 2.0-liter buffered solution created from 1
ID: 512649 • Letter: 1
Question
1. The following question refers to a 2.0-liter buffered solution created from 1.30 M NH3 (Kb = 1.8 × 10–5) and 0.26 M NH4F. When 0.10 mol of H+ ions is added to the solution what is the pH?
5.35
8.56
4.14
9.86
9.85
2. You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 × 10–4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 liter of buffered solution?
286 mL HNO2; 714 mL KNO2
714 mL HNO2; 286 mL KNO2
500 mL of each
587 mL HNO2; 413 mL KNO2
413 mL HNO2; 587 mL KNO2
3. The following question refers to the following system: A 1.0-liter solution contains 0.25 M HF and 1.30 M NaF (Ka for HF is 7.2 × 10–4).
If one adds 0.30 liters of 0.020 M KOH to the solution, what will be the change in pH?
0.73
-0.08
3.87
0.01
-0.19
4. Consider the titration of 300.0 mL of 1.400 M NH3 (Kb = 1.8 × 10–5) with 0.500 M HNO3. After 150.0 mL of 0.500 M HNO3 have been added, the pH of the solution is:
11.92
9.92
4.08
6.92
None of these choices are correct.
Explanation / Answer
1)
NH3 will react with H+ to form NH4+
after reaction,
mol of NH4+ = 0.26 M * 2.0 L + 0.10 mol
mol of NH4+ = 0.620 mol
mol of NH3 = 1.3 M * 2.0 L - 0.10 mol
mol of NH4+ = 2.50 mol
since volume is both in numerator and denominator, we can use mol instead of concentration
Kb = 1.80*10^-5
pKb = - log (Kb)
= - log(1.80*10^-5)
= 4.745
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {0.620/2.50}
= 4.139
use:
pH = 14 - pOH
= 14 - 4.139
= 9.861
pH is 9.861
Answer: 9.86
I am allowed to answer only 1 question at a time
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