In: Chemistry
Need to prepare an acetate buffer solution: 25.00 mL of 0.1 M acetic acid and 0.1 M sodium acetate. The stock acetic acid is 6.0 M. The instructions involve the preparation of a (10-fold) intermediate acetic acid solution. Ka = 1.75 x 10-5 (pKa = 4.76). What are the values for the:
* volume of acetic acid
* grams of sodium acetate (since it's a solid)
* volume of DI water needed
* calculated pH
Thanks.
Acetic Acid: for acetic acid, use the formula N1V1 = N2V2.
As the stock acetic acid is 6.0 M , N1 = 6; V1 = ?
as we need 25 ml of 0.1 M acetic acid , N2 = 0.1 and V2 = 25
Therefore, 6 * V1 = 0.1 * 25, solving for V1 we get V1 = 0.42 ml
Thus for making 25 ml of 0.1M Acetic acid we will need 0.42 ml of 6.0M stock acetic acid and remaining i.e.,24.58ml Dl water.
pH calculation: Ka = 1.75 x 10-5
[H+] = =
pH = -log(0.00000175) = 5.75
Sodium Acetate: Anhydrous sodium acetate has a molar mass of 82.03g.Thus by definition, 82.03g of sodium acetate in 1000ml will yield 1 M solution. Therefore, for a 0.1M solution, you need 8.203g of sodium acetate in 1000ml. Thus for 25 ml of 0.1M sodium acetate you will need 0.2 g (8.203/40) of sodium acetate in 25 ml(1000/40) of Dl water.
pH calculation: Sodium acetate is a weak base, a conjugate base of acetic acid, so:
Ka ×Kb =Kw = 10-14. therefore kb = 10-14/ (1.75 x 10-5) = 5.7142857 x 10-10
[OH-] =
pOH = -log(0.00000755928) = 5.12