1) How many moles of sodium hydroxide would have to be added to 125 mL of a 0.49

Question

1) How many moles of sodium hydroxide would have to be added to 125 mL of a 0.498 M hypochlorous acid solution, in order to prepare a buffer with a pH of 7.180?

2) An aqueous solution contains 0.314 M hydrocyanic acid. How many mL of 0.209 M potassium hydroxide would have to be added to 250 mL of this solution in order to prepare a buffer with a pH of 9.590?

3)An aqueous solution contains 0.424 M ethylamine (C2H5NH2). How many mL of 0.359 M nitric acid would have to be added to 250 mL of this solution in order to prepare a buffer with a pH of 10.600?

Explanation / Answer

SOLUTION:

Dear candidate you have asked 3 different questions. As per guidelines one question is to be solved one time. Here we solve first question.

Q1.

When NaOH is added the following reaction takes place

HOCl + OH- -------> OCl- + H2O

The mixture of OCl- and HOCl at as buffer. The amount of OCl- formed is equal to the amount of NaOH added. Also the amount of HOCl that will react is equal to NaOH added.

pH = pKa + log [OCl-]/[HOCl]

Let x M solution of NaOH is added

pH = pKa + log [x] / [0.498 - x] ; pKa of HOCl = 7.52

7.180 = 7.52 + log [x] / [0.498 - x]

- 0.34 = log [x] / [0.498 - x]

[x] / [0.498 - x] = 0.46

[x] = 0.46 [0.498 - x]

x = 0.23 - 0.46x

1.46x = 0.23

x = 0.16M

Hence molarity of NaOH solution that must be added = 0.16M

Molarity = number of moles / Volume in liters

number of moles = Molarity X Volume in liters

number of moles = 0.16M X 0.125L; Because 125mL = 0.125L

Hence number of moles of NaOH that must be added = 0.02 moles.

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