Question

Determine the pH of solution that results from the addition of 20 ml, 0.01 M Ca(OH)2 to 30 ml of 0.01 M HCl. Answer given is 11.3

Answer 1

Ca(OH)is a strong  base and HCL is a strong acid,therefore it is a strong acid reaction.The reaction will be
H+(aq)+OH-(aq)→H2O(l)
The original no. of moles of H+ in the solution is
30×10^-3 L×0.01M HCL=3×10^-4Mole
The number of moles of OH- added is 20×10^-3L×0.01M HCL=2×10^-4 moles
the resultant moles of H+ is 
3×10^-4-2×10^-4 moles H+(aq)
the total volume of solution=20ml+30ml=50ml=5×10^-3L
(H+)=1×10^-4/5×10^-3=0.02
pH=-log(H+)
=-Log(0.02)
=1.698

Answer 2

Answer:

2.70

Explanation:

The base equivalent are more than acid

Therefore

Normality(N)=(N2V2-N1V1)/V1+V2

N=2×10^-3

Now,

pOH = -log(N)

=2.69

Therefore,

pH = 14 - pOH

=11.30