Question

In: Chemistry

Explain how you would make 500 mL of a 20 mM citric acid buffer at a...

Explain how you would make 500 mL of a 20 mM citric acid buffer

at a pH of 7.0. You must specify the components that you would use

and the masses. Explain what will happen if you add 5 mL of 1.0 M

NaOH to this solution.

Solutions

Expert Solution

pKa1 = 3.13 pKa2 = 4.76 pKa3 = 6.39

if pH goal is 7

then, we must use pKa3, the nearest

HCitrate-2 = citrate-3 + H+

pH = pKa + log(citrate-3/HCitrate-2)

7 = 6.39+ log(citrate-3/HCitrate-2)

(citrate-3/HCitrate-2) = 10^(7 - 6.39) = 4.0738

(citrate-3/HCitrate-2) = 4.0738

total solution molarity:

mmol = MV = (20)(0.5) = 10 mmol

then

(citrate-3/HCitrate-2) = 4.0738 --> citrate-3 = 4.0738 *HCitrate-2

Citrate-3 + HCitrate-2 = 10

4.0738 *HCitrate-2 + HCitrate-2 = 10

HCitrate-2 = 10/(4.0738 +1) = 1.9709

citrate-3 = 4.0738 *HCitrate-2 =1.9709*4.0738 = 8.0290

now..

add:

mol of citrate salt = 10 mmol

Na3C6H5O7

mass = mol*MW = (10*10^-3)(258.06 ) = 2.5806 g of sodium citrate

now, add

HCl at 0.1 M --> mol of acid required = 1.9709

V = mmol/M = 1.9709/0.1 = 19.71 mL of HCl at 0.1 M

then, add enough deionized water up to V = 500 mL mark

this buffer has already pH = 7.

now,

if we add NaOh, then, the pH of the buffer increases drastically, since we will have destroyed this buffer ( plenty of OH will neutralize the acid)

OH- = MV = 5*1 = 5

1.9709-5 = -3.09 mol of OH- leftover

queen_honey_blossom answered 1 year ago

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