What weight of sodium glycolate should be added to 300 mL of 1.00M glycolic acid to produce a buffer solution with a pH of 4.00?
2007-06-19 05:57:32 · 4 answers · asked by ScienceNut 2 in Science & Mathematics ➔ Chemistry
Ka of glycolic acid = 1.48 x 10^-4
pKa = - log 1.48 x 10^-4 =3.83
pH = pK + log [ glycolate ]/ [ glycolic acid]
4.00 = 3.83 + log [ glycolate ] / 1.00
4.00 - 3.83 = 0.17
[ glycolate ] / 1.00 = 10^0.17
[glycolate ] = 1.48 M
1.48 M means 1.48 moles in 1 L; in 300 mL = 0.300 L we have to put 0.444 moles of sodium glycolate.
molar mass = 98.05 g/mol
0.444 x 98.05 = 43.5 g
2007-06-19 06:09:35 · answer #1 · answered by Dr.A 7 · 0⤊ 0⤋
"25mL of 0.10M" ---> this is suited to the acetic acid. pH = pKa + log (base/acid) 5.5 = 4.7447 + log (x / 0.a million) log (x / 0.a million) = 0.7553 x / 0.a million = 5.70 x = 0.570 M (this is the place the previous poster makes a mistake.) MV = mass / molar mass (0.570) (0.0.5) = x / eighty two.03 x = a million.17 g we can assume that dissolving the a million.17 g of sodium acetate into the 25 mL of 0.10 M acetic acid won't strengthen the quantity.
2016-11-06 22:39:53 · answer #2 · answered by Anonymous · 1⤊ 0⤋
pKa of glycolic acid reqd.
2007-06-19 06:03:54 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋
the -log of the Ka is not 3.83
2016-01-17 23:59:55 · answer #4 · answered by Era 1 · 0⤊ 0⤋