Question

A 10.0-ml solution of 0.780 M NH₃ is titrated with a 0.260 M HCL solution. calculate the pH after the following additions of the hcl solution:

A. 0.00

B. 10.0

C. 30.0

D. 40.0

Answer 1

1. At initial we have 0.78 M of NH3 . NH3 reacts with water as follows:

NH3 + H2O -------> NH4+ + OH-

construct ICE table for this

	NH3 + H2O --------->	NH4+      +	OH-
Initial	0.78M	0	0
Change	-x	+x	+x
equilibrium	0.78 -x = 0.78	+x	+x

As , Kb = [NH4+] [OH-]/[NH3]

from text Kb for NH3 = 1.8 X 10^-5 therefore,

1.8 x10^-5 = x²/0.78

x = 1.184 X 10^-5

pOH = -log [OH] = -log[1.184 X10^-5] = 4.92

pH = 14 -4.92 = 9.07

2. On addition of 10ml of HCl

first calculate no. of maoles of NH3 = 0.78 mol/L X 10/100 = 0.0078 moles

moles of HCl = 0.26 mol/L x 10/1000 (L)= 0.0026 moles

Construct ICE table

	NH3 +	HCl ------->	NH4+   +	CL-
Initial	0.0078	0.0026	0	0
Change	-0.0026	-0.0026	0.0026
Equilibrium	0.0052	0	0.0026

use henderson hasblanch equation

pH =pKa + log [NH3]/[NH4+]

for pKa calculated as follows: as we have Kb = 1.8X10^-5

Ka = Kw/Kb = 10^-14/1.8x10-5 = 5.56 X 10^-10
pKa = 9.26

pH = pKa + log [NH3]/[NH4+]
pH = 9.26 + log (0.0056 / 0.0028)
pH = 9.56

3. on adding 30 ml HCl first lets see calculated what is the volume of HCl require to reach equivalent point.

to find the volume of HCL at equivalent point divide not of moles of NH3 by 0.26 M HCL = 0.0076mol/0.26mol/L = 30ml

So 30 ml is the equivalent point.

At equivalent point solution contain only NH+ ion

NH4+ ------> NH3 +H+

Ka = [NH3][H+]/NH4+

from the text Ka for NH4+ = 5.6X10^-10

5.6x 10^-10 = x2/0.0026

x = 0.01456X10^-10

pH = -log[0.01456X10^-10] = 11.83

4. At 40ml HCL , Hcl is the dominant species

10ml of HCL is diluted to 70ml therefore moles of HCl = 0.26 X 70/1000= 0.0182

therefore pH = -log[0.0182] = 1.73

NH4+ --------> NH3 +