How many grams of solid sodium hypochlorite should be added to 1.00 L of a 7.68×

Question

How many grams of solid sodium hypochlorite should be added to 1.00 L of a 7.68×10-2M hypochlorous acid solution to prepare a buffer with a pH of 6.780 ?

Answer in grams sodium hypochlorite =

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How many grams of solid ammonium bromide should be added to 1.00 L of a 5.99×10-2 M ammonia solution to prepare a buffer with a pH of 9.790 ?

Answer in grams ammonium bromide =

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How many grams of solid sodium acetate should be added to 2.00 L of a 7.10×10-2M acetic acid solution to prepare a buffer with a pH of 5.315 ?

Answer in grams sodium acetate =

Explanation / Answer

solution:-

(1) to calculate the pH of a buffer we use the Handerson equation..

pH = Pka + log(base/acid)

Pka for hypochlorous acid is 7.40

moles of hypochlorous acid, HClO = 1.00 L x 7.68 x 10-2mol/L = 7.68 x 10-2 or 0.0768 mol

let's say moles of sodium hypochlorite added = X

let's plug in the values...

6.780 = 7.40 + log(X/0.0768)

6.780 - 7.40 = log(X/0.0768)

-0.620 = log(X/0.0768)

taking antilog..

10-0.620 = (X/0.0768)

0.2399 = (X/0.0768)

X = 0.0184 mol

so, 0.0184 moles of sodium hypochlorite need to be added. let's convert these moles into grams by multiplying the molar mass. molar mass of sodium hypochlorite(NaClO) is 74.44 g/mol.

0.320 mol x 74.44g/mol = 1.37g

Hence, 1.37 g of NaClO are need to be added.

(2)

It could be solved as we solved the first one.

Pkb for NH3 is 4.74. So, Pka would be = 14 - 4.74 = 9.26

moles of ammonia, NH3 = 1.00 L x 5.99 x 10-2mol/L = 5.99 x 10-2mol or 0.0599 mol

Let's say X moles of ammonium bromide, NH4Br are added. Also, NH3 is the base and NH4+ is its conjugate acid. NH4+ comes from NH4Br

let's plug in the values..

9.790 = 9.26 + log(0.0599/X)

9.790 - 9.26 = log(0.0599/X)

0.53 = log(0.0599/X)

taking antilog..

100.53 = (0.0599/X)

3.39 = (0.0599/X)

3.39X = 0.0599

X = 0.0599/3.39

X = 0.0177mol

molar mass of NH4Br is 97/94 g/mol.

so. 0.0177 mol x (97.94 g/mol) = 1.73g

Hence, 1.73 g of NH4Br need to be added.

(3) moles of acetic acid = 2.00 L x (7.10 x 10-2mol/L) = 1.42 x 10-1 mol = 0.142 mol

Let's say X moles of sodium acetate are added.

Pka for acetic acid is 4.74

5.315 = 4.74 + log(X/0.142)

5.315 - 4.74 = log(X/0.142)

0.575 = log(X/0.142)

taking antilog..

3.76 = (X/0.142)

X = 3.76 x 0.142 = 0.534 mol

molar mass of sodium acetate is 82.03 g/mol

so, 0.534 mol x (82.03g/mol) = 43.8 g

so, 43.8 g of sodium acetate need to be added.

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