Question

A 20.00mL of 1.500 M solution of ammonia is being titrated with a 1.00 M solution of HCl. The K_b for ammonia is 1.80×10^-5. Calculate the pH of the solution when the following amounts of HCl have been added.

0.00 mL

10.00 mL

15.00 mL

20.00 mL

30.00 mL

40.00 mL

Answer 1

a) At 0.00 ml

Kb= 1.80x10^-5

C= 1.500M

for weak base

[OH-]= square root of KbxC

[OH-] = square root of ( 1.800x10^-5 x 1.500)

[OH-] = 5.196 x10^-3M

-log[OH-] = -log(5.196x10^-3]

POH= 2.28

PH+POH=14

PH=14-POH

PH= 14-2.28

PH= 11.72

b) At 10.00ml

NH3= 20.00 ml of 1.500M

number of moles of NH3 = 1.500M x 0.0200L= 0.03mole

number of moles of NH3= 0.03 moles

HCl= 10.00ml of 1.00M

number of moles of HCl= 1.0M x 0.01L = 0.01 mole

number of molesof HCl= 0.01 moles

NH3 + HCl -------------------- NH4Cl

0.03       0.01                      0

-0.01     - 0.01                      +0.01

0.02         0                           + 0.01

after addition of HCl,

remaining number of moles of NH3 = 0.02 mole

number of moles of NH4Cl= 0.01 mole

Kb= 1.8x10^-5

-log[Kb] = -log[1.8x10^=5]

PKb= 4.74

POH= PKb + log[salt]/[base]

POh= 4.74 + log(0.01/0.02)

POH= 4.44

PH= 14- POH

PH= 14 - 4.44= 9.56

PH= 9.56

c) At 15 ml

HCl= 15.00ml 0f 1.00M

number of moles of HCl= 1.00M x0.015 L= 0.015 mole

after additon of Acid ,

remaining number of moles of NH3 = 0.03 - 0.015= 0.015 moles

number of moles of salt = 0.015 moles

so it is the half-equivalent point or mid point

at half equivalent point

POH= PKb

POH= 4.74

PH- 14-4.74= 9.26

PH= 9.26

d)

At 20.00ml

Number of moles of HCl= 1.0M x0.0200L= 0.02 moles

after addion of HCl

remaining number of moles of NH3 = 0.03 - 0.02= 0.01 moles

number of moles of salt = 0.02 mole

POH= 4.74 + log(0.02/0.01)

POH= 5.04

PH= 14-5.04

PH= 8.96

e)At 30.00ml

number of moles of HCl= 1.0Mx0.0300L= 0.03 moles

so it is the equivalent point

at equivalent point

PH= 7 - 1/2[PKb + log C]

C= number of moles of salt or base/total volume                total volume = 20.00+30.00= 50.00ml= 0.05L

C= 0.03 /0.05=0.6M

PH= 7 - 1/2[4.74 + log(0.6)]

PH= 4.74

f) At 40.00ml

Number of moles of HCl= 1.0M x0.0400L= 0.04 moles

number of moles of HCl is more than that of base. so there is no possibility of buffer.

remaining number of moles of HCl= 0.04 - 0.03 = 0.01 moles

total voume = 20.00 + 40.00= 60.00 ml= 0.06L

[H+]= 0.01/0.06=0.167

[H+] = 0.167M

-log[H+] = -log(0.167)

PH= 0.78

PH= 0.78