Calculation of the Concentrations of Various Species in Salt Solution (Coursework Sample)

Report sheet
Part A: Determination of Ka and Kb
Ka of NH4Cl ⇒Ka=H3O+[A-][HA]=x*x0.1-x=x20.1-x
H3O+=10-PH=10-6.01=9.77×10-7=x
Ka=(9.77×10-7)20.1-9.77×10-7=9.54538×10-12
Ka-Kb=Kw=1×10-14
Kb=1×10-149.54538×10-12=0.00104=1.04×10-3
Kb of Na2CO3⇒Ka=H3O+[A-][HA]=x*x0.1-x=x20.1-x
pOH=14-pH=14-10.93=3.07
OH-=10-POH=10-3.07=8.511×10-4=x
Kb=(8.511×10-4)20.1-8.511×10-4=7.30589×10-6
Ka=1×10-147.30589×10-6=1.36876×10-9
Ka of Al (NO3)3 ⇒Ka=H3O+[A-][HA]=x*x0.1-x=x20.1-x
H3O+=10-PH=10-3.45=3.548×10-4=x
Ka=(3.548×10-4)20.1-3.548×10-4=1.2633×10-6
Kb=1×10-141.2633×10-6=7.91578×10-9
Solution

Hydrolysis Net Ionic equation

Ka or Kb expression

Value of Ka or Kb

NH4Cl

NH4+H2O⇋NH3+H3O+

Kb=NH4+[H3O+][NH4]

Ka=9.54538×10-12

Na2CO3

CO32-+H2O⇋HCO3-+OH-

Kb=HCO3-[OH-][CO32-]

Ka=7.30589×10-6

Al (NO3)3

Al3++H2O⇋AlOH2++H

Kb=AlOH[H+][Al3+]

Ka=1.2633×10-6

NaOAc

NaOAc+H2O⇌CH3COO-+NaOH

Kb=CH3COO-[NaOH][NaOAc]

Ka = 1.613*10-5

Solution

Measured pH

[H+]

[OH-]

200 mL H2O

6.29

5.12861×10-7

0.09999948714

200 mL H2O-boiled

7.00

0.0000001

0.0999999

25 mL H2O

6.29

5.12861×10-7

0.09999948714

0.1 M NH4Cl

6.01

9.77237×10-7

0.09999902276

0.1 M Na2CO3

10.93

1.1749×10-11

0.09999999999999

0.1 M Al(NO3)3

3.45

3.548×10-4

0.0996452

Part B: Buffer Capacity
The mass of NaOAc.3H2O
Moles of NaOAc.3H2O
3.374320=0.01 mol
Volume of the buffer
22.4 L ×0.01 mol=0.224 L
Buffer=Number of moles of H+/OH-Change in pH(Volume of the buffer)
BC for buffer=10.2×0.224=22.32
pH=4-log22.32=3.37
BC for buffer+NaOH=1(4.72-4.51)×22.4=0.21258
pH=4--log0.21258=0.67
BC for H2O+HCl=1(7-0.95)×22.4=0.00737
pH=-log0.00816=2.13
BC for H2O+NaOH=1(12.47-7)×22.4=0.00816
pH=14+log0.00816=11.91
Solution

Measured pH

Calculated pH

Buffer solution

4.51

4.73

6.0 M HCl

-0.50

0.78

6.M NaOH

13.64

13.22

Buffer + HCl

4.00

4.67

Buffer + NaOH

4.72

3.37

H2O + HCl

0.95

2.13

H2O+ NaOH

12.47

11.911

Question 1
Solution of sodium acetate will be basic. Sodium acetate is the salt formed by the acetic acid (weak acid) and sodium hydroxide (strong base). Solution of strong base and weak acid is always basic.
CH3-COOH+NaOH⇌CH3-COONa+H2O
Question 2
Solution of ammonium chloride formed by ammonia (weak base) and hydrochloric acid (a strong acid). A solution of a strong acid and weak base is always acidic. So the solution of NH4Cl will be an acidic
Question 3
Ka of HCN = 6.2*10-10
use:Kb =KwKa
Kw is dissociation constant of water whose value is 1.0*10-14 at 25oC
Kb =1.0*10-14Ka
Kb =1.0*10-146.2*10-10
Kb = 1.613*10-5
CN- dissociates as CN- + H2O → HCN + OH-
0.1 0 0
0.1-x x x
Kb =HCNOH-CN- ⇒Kb = x2(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x2c; so, x =(Kb*c)
x = ((1.613*10^-5)*0.1) = 1.27×10-3
Since x is comparable c, our assumption is not correct
We need to solve this using Quadratic equation
Kb = x2c-x
1.613*10-5=x20.1-x
1.613*10-6 - 1.613*10-5*x = x2
x2+ 1.613*10-5 *x-1.613*10-6 = 0
x =-b + sqrtb2-4*a*c2a
x = 1.262*10-3 and x = -1.278*10-3
since x can't be negative, the possible value of x is
x = 1.262*10-3
use:
pOH = -log [OH-]= -log (1.262*10-3)= 2.899
PH = 14 - pOH = 14 - 2.899= 11.1
Question 4
* HCl is a strong acid and it will dissociate completely.
mmol of HCl present = Molarity × Volume ml= 1.00 M ×1.7ml= 1.7 mmol
Since HCl is acid, it reacts with the conjugate base (HPO42- ) present in the buffer and the amount of HPO42- decreases and that of the conjugate base increases.
HPO42-+H3O+→H2PO42-
1 mmol H3O+ react 1 mmol HPO42-
1.7 mmol H3O+ react 1.7 mmol HPO42-
After addition
mmol HPO42-=18.2 mmol-1.7 mmol=16.5 mmol
mmol H2PO4-=11.8 mmol-1.7 mmol=13.5 mmol
So the molar quantities are H3O+=1.7 mmol, HPO42-=16.5 mmol, H2PO4-=13.5 mmol
* pH = pKa + logHPO42-H2PO4-
6.83=6.64 + logHPO42-H2PO4-
[HPO42-]=1.55 [H2PO4-]
[H2PO4-]+[HP