HW Set #3 24 pt CHE 120 DO ALL WORK ON A SEPARATE SHEET SHow woRK (No work, No c
ID: 497234 • Letter: H
Question
HW Set #3 24 pt CHE 120 DO ALL WORK ON A SEPARATE SHEET SHow woRK (No work, No credit use constant info in book or tables from class (4pt) 1. For the following hypothetical reaction that begins with only A and B, where, initially the concentrations of [A] 0.100 M the concentration of (Bl 0.200 M Arg) 2 B 3 ceo, 4 Day Ko 1.00 x 1002 What is the concentration of every species at equilibrium? (8pt) 2. A 200 mL solution of 0.100 M solution of HF is slowly mixed with 0.1 M NaOH. Determine the pH after addition of o ml of NaOH, 50 L of NaOH, 100 ml of aOH, 200 mL of NaOH, 250 mL of NaoH (8pt) 3. A 200 mL solution of 0.100 M solution of Ammonia, NH is slowly mixed with 0.100 M HCI. Determine the pH after addition of o mb of HCI, 50 ml of HCI, 1 mL of HCI, 200 mL of HCI, 250 mL of HCl (4pt) Describe in exact detail (including gram quantities and identity of species) how you would 4. create a 500 mL solution of buffer with a pH 8.0 20Explanation / Answer
Q2.
First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well
M = moles / volume
pH = -log[H+]
a) HA -> H+ + A-
Ka = [H+][A-]/[HA]
a) no NaOH
Ka = [H+][A-]/[HA]
Assume [H+] = [A-] = x
[HA] = M – x
Substitute in Ka
Ka = [H+][A-]/[HA]
Ka = x*x/(M-x)
7.2*10^-4 = x*x/(0.1-x)
This is quadratic equation
x =0.00813
For pH
pH = -log(H+)
pH =-log(0.00813)
pH in a = 2.08990
b) 50 ml KOH
This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)
Use Henderson-Hassebach equations!
NOTE that in this case [A-] < [HA]; Expect pH lower than pKa
pH = pKa + log (A-/HA)
initially
mmol of acid = MV = 200*0.1 = 20 mmol of acid
mmol of base = MV = 0.1*50 = 5 mmol of base
then, they neutralize and form conjugate base:
mmol of acid left = 20-5 = 15 mmol
mmol of conjguate left = 0 + 5 = 5
Get pKa
pKa = -log(Ka)
pKa = -log(7.2*10^-4) = 3.142
Apply equation
pH = pKa + log ([A-]/[HA]) =
pH = 3.142+ log (5/15) = 2.664
c) for 100 ml
This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)
Use Henderson-Hassebach equations!
NOTE that in this case [A-] < [HA]; Expect pH lower than pKa
pH = pKa + log (A-/HA)
initially
mmol of acid = MV = 200*0.1 = 20 mmol of acid
mmol of base = MV = 0.1*100 = 10 mmol of base
then, they neutralize and form conjugate base:
mmol of acid left = 20-10 = 10 mmol
mmol of conjguate left = 0 + 10= 10
Get pKa
pKa = -log(Ka)
pKa = -log(7.2*10^-4) = 3.142
Apply equation
pH = pKa + log ([A-]/[HA]) =
pH = 3.142+ log (10/15) = 3.142
d) Addition of Same quantitie of Acid/Base
This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium
We will need Kb
Ka*Kb = Kw
And Kw = 10^-14 always at 25°C for water so:
Kb = Kw/Ka = (10^-14)/(7.2*10^-4) = 1.388*10^-11
Now, proceed to calculate the equilibrium
H2O + A- <-> OH- + HA
Then K equilibrium is given as:
Kb = [HA][OH-]/[A-]
Assume [HA] = [OH-] = x
[A-] = M – x
Substitute in Kb
1.388*10^-11 = [x^2]/[M-x]
recalculate M:
mmol of conjugate = 20 mmol
Total V = V1+V2 = 200+200 = 400
[M] = 20 mmol/400 = 0.05 M
1.388*10^-11 = [x^2]/[0.05 -x]
x = 8.33*10^-7
[OH-] =8.33*10^-7
Get pOH
pOH = -log(OH-)
pOH = -log (8.33*10^-7) = 6.079
pH = 14-pOH = 14-6.079= 7.92
e) Addition of base
There will be finally an Excess of Base!
mol of acid < mol of base
Calculate pOH directly
[OH-] = M*V / Vt
mmol of acid = MV = 200*0.10 = 20
mmol of base = MV = 250*0.1 = 25
therefre,
mmol of strng base left = 25-20 = 5 mmol
Vtotal = 250+200 = 450 mL
[OH-] = 5/450 = 0.01111
pOH = -log(OH-)
pOH = -log(0.01111) = 1.954
pH = 14-pOH = 14-1.954= 12.046
pH = 12.046
For Q3.
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