Question

A 50.0 mL solution of 0.200 M ammonia (pKa of ammonia is 9.24) is titrated by 0.100 M HCl. Calculate the pH of the solution when, A.) the volume of HCl is 20.00 mL B.) the volume of HCl is 100.0 mL

Answer 1

a)

NH3 is a weak base. pKb = 14 - pKa = 14 - 9.24 = 4.76

Also, pKa = 9.24 so, Ka = 10^-pKa = 10^-9.24

Initial millimoles of NH3 = molarity*volume in ml = 0.2M*50ml = 10 millimoles

millimoles of HCl added = molarity*volume in ml = 0.1M*20ml = 2 millimoles

If we add 20 ml 0.1 M HCl the following reaction will take place

NH3 + HCl -> NH4Cl

So, each mole of NH3 reacts with 1 mole of HCl to produce 1 mole of NH4Cl

So, 2 millimoles of HCl will react with 2 millimoles of NH3 to produce 2 millimoles of NH4Cl

Finally we will have  2 millimoles of NH4Cl and 10 millimoles - 2 millimoles = 8 millimoles of NH3

Mixture of weak base NH3 and its salt with strong acid NH4Cl is a buffer solution.

pOH of a buffer solution is given by henderson hasselbalch equation

pOH = pKb + log(salt/base) where salt/base is the ratio of millimoles of salt and base in our case salt/base = 2 millimoles/8 millimoles = 1/4

so, pOH = 4.76 + log(1/4) = 4.158

so, pH = 14 - pOH = 14- 4.158 = 9.84

b)

If 100ml of 0.1M HCl is added the reaction with NH3 will go to completion and thus finally we will have 10 millimoles of NH4Cl in a total volume of 50ml+100ml = 150ml so, its concentration is millimoles/volume in ml = 10 millimoles/150 ml = 0.0667 M

NH4Cl is a salt of weak base and strong acid and thus go under salt hydrolysis as -

NH4Cl -> NH4+ + Cl-

NH4+ -> NH3 + H+

I 0.0667 0 0

C -0.0667x +0.0667x +0.667x

E 0.0667-0.0667x 0.0667x 0.0667x

Ka = [H+][NH3]/[NH4+] = 0.0667x*0.0667x/(0.0667-0.0667x) = 10^-9.24

0.0667x²/(1-x) = 10^-9.24 = 5.75*10^-10

Assuming x<<1 and thus neglecting it compared to 1 we get

0.0667x² = 5.75*10^-10

x = sqrt( 5.75*10^-10/0.0667) = 9.288*10^-5

so, H+ = 0.0667x = 0.0667*9.288*10^-5

= 0.619*10^-5 M

so, pH = -log(H+) = -log( 0.619*10^-5)

pH = 5.2