2006-09-23 03:16:02 · 3 answers · asked by joy 1 in Science & Mathematics ➔ Chemistry
CH3COOH + H2O <====> CH3COO- + H3O+
the pKa of CH3COOH is 4.75
using Henderson-Hasselbach Eqn
pKa=pH - log([CH3COO-]/[Ch3COOH])
at pH = 2.00
4.75=2.00-log([CH3COO-]/[Ch3COOH])
2.75= -log([CH3COO-]/[Ch3COOH])
antilog(-2.75) = [CH3COO-]/[Ch3COOH]
the ratio then is
[CH3COO-]/[Ch3COOH] = 1.78 x 10^-3...or
[CH3COOH]/[Ch3COO-] = 561.8
the first value is when CH3COO- is in the numerator
the 2nd value is when CH3COOH is in the numerator
this just tells that the moles of acetic acid is relatively high compared to the acetate ion, i.e. 562:1...
at pH = 3.00
4.75=3.00-log([CH3COO-]/[Ch3COOH])
1.75= -log([CH3COO-]/[Ch3COOH])
antilog(-1.75) = [CH3COO-]/[Ch3COOH]
the ratio then is
[CH3COO-]/[Ch3COOH] = 0.0178...or
[CH3COOH]/[Ch3COO-] = 56.18
the first value is when CH3COO- is in the numerator
the 2nd value is when CH3COOH is in the numerator
this just tells that the moles of acetic acid is still relatively high compared to the acetate ion, i.e. 56:1...
best answer?
2006-09-23 04:35:43 · answer #1 · answered by teroy 4 · 0⤊ 0⤋
the pH is the exponent of the concentration of hydrogen ions
pH 2 = 10^2
pH 3 = 10^3
there for pH2 contains 10 times the number of hydrogen ions than pH 3
2006-09-23 03:22:16 · answer #2 · answered by The Cheminator 5 · 0⤊ 0⤋
01.ionization constant of acid is
Ka=[CH3COO-][H30+]/[CH3COOH]=0.000018
02.H3O+ concentration is antilog of ph=.01
03.solving eq. 01 [CH3COOH]/CH3COO-]=[H3O+]/Ka
=.01/.000018
=555.55
04.like this another ratio is =.001/.000018=55.55
2006-09-23 05:22:20 · answer #3 · answered by shakti s 1 · 0⤊ 0⤋