Need to prepare an acetate buffer solution: 25.00 mL of 0.1 M acetic acid and 0.1...

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. K_a = 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.

Solutions

Expert Solution

Acetic Acid: for acetic acid, use the formula N₁V₁ = N₂V_2.

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 =K_w = 10^-14. therefore kb = 10^-14/ (1.75 x 10^-5) = 5.7142857 x 10^-10

[OH^-] =

pOH = -log(0.00000755928) = 5.12

queen_honey_blossom answered 1 year ago

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