You are given 60 mL of 0.50 M acetic acid/ acetate buffer to test. The starting
ID: 878527 • Letter: Y
Question
You are given 60 mL of 0.50 M acetic acid/ acetate buffer to test. The starting composition of the two major species are:
Concentration of CH3COOH: 0.300 M
Concentration of CH3COO-: 0.200 M
A. Use the henderson hasselbach equation to estimate the initial pH of the buffer.
B. You add 1.0 mL of 1.00 M HCl to the buffer. Calculate the molarity of H3O+ added as HCl, and the final molarities of acetic acid and acetate ion at equilibrium. What is the new value of the pH?
C. Now take a fresh 60 mL of the buffer and add 1.0 mL of 1.00 M NaOH. Using steps similar to those above calculate the new pH of the solution.
Please show your work. Thank you.
Explanation / Answer
Solution:
Given-
[CH3COOH] = 0.300 M
[CH3COO-] = 0.200 M
Volume = 60 mL = 0.06 L
Pka of CH3COOH = 4.756
Dissociation of CH3COOH is as follows
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
A.
sing the henderson hasselbach equation
PH = PKa + log10 ([CH3COO-]/[CH3COOH])
= 4.756 + log10 ([0.200 M]/[0.300 M]) = 4.756 + log10 (0.666)
= 4.756 – 0.1760
PH = 4.58
Initial pH of the buffer = 4.58
B. You add 1.0 mL of 1.00 M HCl to the buffer. Calculate the molarity of H3O+ added as HCl, and the final molarities of acetic acid and acetate ion at equilibrium. What is the new value of the pH?
Given-
Volume = 1.0 mL = 0.001 L
Molarity of HCl = 1.0 M
By using volume and molarity we get moles of HCl
#moles of HCl = Volume in L X molarity =0.001 L X 1.0M
= 0.001 M
Given-
[CH3COOH] = 0.300 M
Volume = 60 mL = 0.06 L
By using volume and molarity we get moles of acetic acid
#moles of acetic acid = Volume in L * molarity
= 0.06 L * 0.300 M
= 0.018 mol acetic acid.
Molarity of acetic acid = # moles / volume in L
= 0.018mol acetic acid / 0.0600 L ( here we consider total volume of buffer)
= 0.3 M
For Acetic acid :
CH3COOH (aq) + H2O ( l ) à CH3COO- (aq) + H3O + (aq)
I 0.3 M 0 0
C -x + x +x
E ( 0.3-x) x x
Ka expression
Ka = [ CH3COO- (aq)][ H3O + (aq)] / [ CH3COOH (aq) ]
Ka for acetic acid = 1.8 * 10^-5
1.8 * 10^-5 = x^2 / (0.3 M-x)
By using 5% approximation we neglect x at the denominator.
1.8 * 10-5 = x2 / (0.3 M)
( 1.8 * 10-5 ) X 0.3 = x^2
X2 = 0.54 x 10-5 M
X = 0.00232 M
[H3O+] = x = 0.00232 M
We add both moles of H3O from HCl and acetic acid
Total moles = 0.001 mol HCl + 0.00232 mol H3O+
=0.00332 mol H3O+
Now concentration of H3O = #mol H3O / volume in L
= 0.00332 mol H3O+ / 0.06 L
= 0.0555 M
PH = -log[H3O] = -log[0.00555] = 1.255
PH of CH3COOH = 1.255
C. Now take a fresh 60 mL of the buffer and add 1.0 mL of 1.00 M NaOH. Using steps similar to those above calculate the new pH of the solution.
Given-
Molarity = 1.0 M NaOH
Volume = 1.0 mL = 0.001 L
By using volume and molarity we get moles of acetic acid
#moles of NaOH = Volume in L * molarity
= 0.01 L * 0.5 M
= 0.01 mol NaOH
Molarity of NaOH = # moles / volume in L
= 0.01mol NaOH / 0.0600 L ( here we consider total volume of buffer)
= 0.166 M
We use ICE chart
For NaOH :
NaOH(s) + H2O(l) à Na+ + OH- + H20
I 0.166 M 0 0
C -x + x +x
E ( 0.166-x) x x
Ka expression
Kb = [OH- (aq)][ Na + (aq)] / [NaOH (aq) ]
Ka for NaOH = 0.2
0.2 = x^2 / (0.166 M-x)
By using 5% approximation we neglect x at the denominator.
0.2 = x2 / (0.166 M)
0.2 X 0.166 = x^2
X2 = 0.332 M
X = 0.576 M
[OH-]= x = 0.576 M
Lets write the reaction between NaOH and CH3COOH
Ch3COOH + Na+OH- CH3COO-Na+ + H2O
Lets calculate number of of CH3COOH
#moles of CH3COOH = Volume in L * molarity ( here we consider total volume of buffer)
= 0.06 L X 0.5 M
= 0.03 mol
CH3COOH (aq) + H2O ( l ) à CH3COO- (aq) + H3O + (aq)
I 0.03 M 0 0
C -x + x +x
E ( 0.03-x) x x
Ka expression
Ka = [ CH3COO- (aq)][ H3O + (aq)] / [ CH3COOH (aq) ]
Ka for acetic acid = 1.8 * 10^-5
1.8 * 10^-5 = x^2 / (0.03 M-x)
By using 5% approximation we neglect x at the denominator.
1.8 * 10-5 = x2 / (0.03 M)
( 1.8 * 10-5 ) X 0.03 = x^2
X2 = 0.054 x 10-5 M
X = 0.000734 M
[H3O+] = x = 0.000734 M
To find the concentration of CH3COO-
We use the Henderson-Hasselbalch Equation
PH = PKa + log [Base]/[Acid]
= 4.756 + log [0.576]/[0.000734]
PH = 2.89
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.