What mass of solid NaCH3CO2 (molar mass = 82.0 g/mol) should be added to 1.00 L of 0.350 M CH3CO2H to make a buffer of pH = 5.20?
(Ka for CH3CO2H = 1.8×10-5)
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answers:
pisgahchemist say: Buffer solution .....

A buffer is a solution of a weak acid and its conjugate base. The weak acid acetic acid (aka ethanoic acid) and the conjugate base is the acetate ion, CH3COO-.

pH = -log[H+]
[H+] = 10^-pH = 10^-5.2 = 6.31x10^-6

CH3COOH(aq) <==> H+ + CH3COO-
0.350M ................ 6.31x10^-6 ...x

Ka = [H+][CH3COO-] / [CH3COOH]
1.78x10^-5 = 6.31x10^-6(x) / 0.350
x = 0.987

The acetate ion concentration is 0.987 mol/L. Therefore, 0.987 mol of sodium acetate must be dissolved in 1.00L of solution, assuming the volume does not change.

0.987 mol CH3COONa x (82.0g CH3COONa / 1 mol CH3COONa) = 81.0g CH3COONa
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Fern say: CH3COOH ⇐⇒ H+ + CHCOO-

Ka =[H+][CH3COO-] / [CH3COOH]

pH = 5.20 ; [H+] = 10^-5.20 = 6.31 x 10^-6

6.31 x 10^-6 [CH3COO-] / 0.350 = 1.8 x 10^-5

[CH3COO-] =1.8 x 10^-5(0.350)/6.31 x 10^-6

[CH3COO-] = 0.998 M

NaCH3COO =⇒ Na+ + CH3COO-
NaCH3COO = 82.0 grams/mole

0.998 moles NaCH3CO2 x 82.0 grams NaCH3CO2/mole NaCH3CO2 = 82.0 grams
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