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
Answers
Answered by
15
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
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
Answered by
9
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
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