1. How many moles of sodium hydroxide would have to be added to 125 mL of a 0.49
ID: 969601 • Letter: 1
Question
1. How many moles of sodium hydroxide would have to be added to 125 mL of a 0.498 M nitrous acid solution, in order to prepare a buffer with a pH of 3.100?
2. An aqueous solution contains 0.328 M hydrofluoric acid.
How many mL of 0.388 M potassium hydroxide would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 3.610?
3. Determine the pH during the titration of 29.4 mL of 0.107 M hydrobromic acid by 0.300 M sodium hydroxide at the following points:
(1) Before the addition of any sodium hydroxide:
(2) After the addition of 5.25 mL of sodium hydroxide:
(3) At the equivalence point:
(4) After adding 13.4 mL of sodium hydroxide:
4. Determine the pH during the titration of 16.3 mL of 0.108 M nitric acid by 2.32×10-2 M barium hydroxide at the following points:
(1) Before the addition of any barium hydroxide:
(2) After the addition of 19.0 mL of barium hydroxide:
(3) At the equivalence point:
(4) After adding 47.8 mL of barium hydroxide:
5. When a 17.6 mL sample of a 0.434 M aqueous acetic acid solution is titrated with a 0.303 M aqueous
sodium hydroxide solution, what is the pH after 37.8 mL of sodium hydroxide have been added?
pH =
6. A 30.4 mL sample of a 0.457 M aqueous hypochlorous acid solution is titrated with a 0.479 M aqueous
sodium hydroxide solution. What is the pH after 11.3 mL of base have been added?
pH =
Explanation / Answer
1. No of mol of HNO2 = 125/1000*0.498 = 0.06225 mol
pka of HNO2 = 3.15
pH = pka + log(salt/acid)
3.1 = 3.15+log(x/(0.06225-x))
x = 0.02933 mol
No of mol of NaOH required = 0.02933 mol
2. pka of HF = 3.17
No of mol of HF = 125/1000*0.388 = 0.0485 mol
pH = pka + log(salt/acid)
3.61 = 3.17+log(x/(0.0485-x))
x = 0.0356 mol
volume of KOH required = 0.0356/0.388*1000 = 91.7 ml
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.