(1) Preparation of acetate buffer: Calculate the amount of sodium acetate trihyd

Question

(1) Preparation of acetate buffer:

Calculate the amount of sodium acetate trihydrate needed to make 1.00 L of a

0.0100 M acetate buffer at pH 5.5

(2) Preparation of 100 ppm ammonium chloride solution:

Calculate the grams of ammonium chloride (NH4Cl) required to make 250 ml of a solution that is 100 mg/L (or parts per million, ppm) NH4+ as N. (There are 14.0g N/mol NH4+.) In lab we will be making solutions that are in ppm N, not ppm NH4+. This is because the NH4+ can become NH3 or environmentally converted to NO3-. These are 1:1:1 processes so the ppm of N stays constant even though the form may change.

Explanation / Answer

Solution (1) :- From Handerson equation,

pH = pKa + log(base/acid)

Total moles of buffer = 1.00 L x (0.0100 mol/L) = 0.0100 mol

Let's say X moles of sodium acetate trihydrate are added. Then moles of acetic acid would be 0.0100 - X.

pKa for acetic acid is 4.74. Let's plug in the values in the Handerson equation,

5.5 = 4.74 + log[X/(0.0100 - X)]

5.5 - 4.74 = log[X/(0.0100 - X)]

0.76 = log[X/(0.0100 - X)]

Taking antilog..

100.76 = X/(0.0100 - X)

5.75 = X/(0.0100 - X)

On cross multiply...

5.75(0.0100 - X) = X

0.0575 - 5.75X = X

0.0575 = X + 5.75X

0.0575 = 6.75X

X = 0.0575/6.75

X = 0.00852 mol

Sodium acetate trihydrate is NaC2H3O2*3H2O. It's molar mass is 136.09 g/mol.

0.00852 mol x (136.09 g/mol) = 1.16 g

So, 1.16 g of sodium acetate trihydrate are added.

(2):- 250 mL x (1L/1000mL) x (100 mg of N/L) x (1g/1000mg) x (1mol NH4Cl/14.0 g N) x (53.49 g/1mol) x (1000mg/1g)   = 95.5 mg NH4Cl

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