Why does a buffer made of equal concentrations of CH3COOH and CH3COO- not change in pH when a small amount of NaOH (a strong base) is added to the solution?
2007-04-16 11:00:03 · 3 answers · asked by theweirdguy1 2 in Science & Mathematics ➔ Chemistry
The Ch3COO- competes with the Na for the -OH instead of forming H2O and Ch3COONa.
2007-04-16 11:05:42 · answer #1 · answered by hunkra2001 2 · 0⤊ 0⤋
Weird guy, that's a good question. It's all in the math. Lets try a mind experiment and create a 1 liter solution in which we have put 0.5 moles of the acid and the acetate. Then the equilibrium expression says:
[HOAc][OH-]/[OAc-] = Kw/Ka. We know Kw is 1x10-14 and Ka = 1.8x10-5 (doesn't everyone know that!!). So we have:
0.5 [OH-] / 0.5 = 10x10-15/1.8x10-5= 5.5x10-10
Let's add 0.1 moles of NaOH to this without changing volume appreciably. It reacts with HOAc to form more acetate ion. Then the [OH-] is given by: 0.4 [OH-]/0.6 = 5.5x10-10, and
[OH-] = 8.2 x 10-10
Had there been no buffer, the [OH-] would have become 0.1 mole/liter.
That's how the buffer buffers. Have a nice day.
2007-04-16 11:15:55 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
3-methylcyclohexene There are six carbons interior the ring, which is composed of one double bond, making it a cyclohexene. The double bond carbons are given the numbers a million and a pair of by skill of default (no longer mandatory in call), so the methyl team is on the 0.33 carbon. positioned that all and sundry jointly and you get the IUPAC call.
2016-12-10 03:44:31 · answer #3 · answered by eatough 4 · 0⤊ 0⤋