solution tested pH [H=}(M) [C2H3O2-](M) K Percentdissociation 0.10 HC2H3O2 2.85
ID: 852983 • Letter: S
Question
solution tested pH [H=}(M) [C2H3O2-](M) K Percentdissociation
0.10 HC2H3O2 2.85
0.010 HC2H3O2 3.19
0.0010 M HC2H3O2 3.58
1. Write the net ionic equation describing the dissociation ofacetic acid in aqueous solution.
2. Calculate [H+] and[ C2H3O2-] for each concentration above, showing a sample calculation here.
3. Write the equilibrium constant expression for acetic acid, andcalculate a K value and an apparent percent dissociation for eachconcentration showing sample calculations.
4. What is the affect of dilution on the percent dissociation? why should this behavior be expected?
Explanation / Answer
. Write a balanced net ionic equation to explain the observed pH for
it looks like only acetic acid was tested:
HC2H3O2 (aq) --> H+ (aq) & C2H3O2- (aq)
======================================...
2.how do your observations indicate that the buffer solution maintains near constancy of pH?
even through the [HC2H3O2] @ 0.010 molar was made 10X's weaker @ 0.0010 Molar & also made 10X's stronger @ 0.1 Molar the pH indicated that the [H+] did not change 10X's times ,(which would have been from 3.19 to 4.19 , or from 3.19 to 2.19...
it hardly changed at all, only from 3.19 to 3.58 & from 3.19 to 2.85
======================================...
3. How do your observations indicate that a buffer solution has a finite capacity?
its moles are altered non the less,
so if a person were to continue to stress a buffer it would run out
the answer to question # 3 below shows that the acid went form ionizing 1% to 6% to 26 % as the exp progressed
=====================================
4.Why did the color of Methyl Orange change at all in the presence of the buffer components?
indicators need a change in two full pH units to do a color change, when the start to change. methyl red is red @ about pH of 4.5 & below... it changes to full yellow @ pH of 6...
all of your pH's were at levels below 4.5 where it is red, red, red
======================================
I would like to change your chart based on...
acetic acid does not turn lose all of its H+'s & C2H3O-'s ....
usually it turns loose only about 1% of it's ions before someone starts shifting its equilibrium around:
by answering this next question:
2. Calculate [H+] and[ C2H3O2-] for each concentration above, by doing a "10^-x" with the x = pH ..... and realizing the equation above relates that H+ & C2H3O2- are released in a 1:1 ratio
solution tested .... .... pH ... 10^-pH
0.10 HC2H3O2 .. .... 2.85 [H+] & [ C2H3O2-] = [1.4e-3M]
0.010 HC2H3O2 .. .. 3.19 [H+] & [ C2H3O2-] = [6.5e-4M]
0.0010 M HC2H3O2 .. 3.58 [H+] & [ C2H3O2-] = 2.6e-4M]
3. Write the equilibrium constant expression for acetic acid, and calculate a K value
0.10 HC2H3O2 .. [H+] & [ C2H3O2-] = [1.4e-3M]
K = [H+] [C2H3O2-] / [HC2H3O2]
K = [1.4e-3M] [1.4e-3M] / [0.10]
K = 2.0e-5
==========
0.010 HC2H3O2 .. [H+] & [ C2H3O2-] = [6.5e-4M]
K = [H+] [C2H3O2-] / [HC2H3O2]
K = [6.5e-4M] [6.5e-4M] / [0.010]
K = 4.2e-5
==========
0.0010 M HC2H3O2 .. [H+] & [ C2H3O2-] = [2.6e-4M]
K = [H+] [C2H3O2-] / [HC2H3O2]
K = [2.6e-4M] [2.6e-4M] / [0.0010]
K = 6.9 e-5
==================
and an apparent percent dissociation for each concentration showing sample calculations.
0.10 HC2H3O2 .. .... 2.85 [H+] & [ C2H3O2-] = [1.4e-3M]
% =( [1.4e-3M] / 0.10) times 100 = 1.4%
0.010 HC2H3O2 .. .. 3.19 [H+] & [ C2H3O2-] = [6.5e-4M]
% =( [6.5e-4M] / 0.010) times 100 = 6.5 %
0.0010 M HC2H3O2 .. 3.58 [H+] & [ C2H3O2-] = 2.6e-4M]
% =( [2.6e-4M] / 0.0010) times 100 = 26 %
======================================...
4. What is the affect of dilution on the percent dissociation?
as it got more dilute ,it ionizes more
why should this behavior be expected?
as more & more water was present , more & more molecules of acetic acid were split into ions by the water
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.