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when 20ml of 0.1N NaOH is mixed with 20ml of 0.05M Ca(OH)2 at 25C
Equivalent weight of NaOH = molecular weight of NaOH because its acidity is one
NORMALITY of NaOH = MOLARITY of NaOH =.1
Milli mole of NaOH=molarity x volume = 20x.1 = 2
NaOH——->Na+ + OH-
2 milli mole of NaOH produce 2 milli mole of OH-
Milli mole of Ca(OH)2=molarity x volume = 20x.05 = 1
Ca(OH)2———→Ca+2 + 2OH-
1 milli mole of Ca(OH)2 produce 2 milli mole of OH-
Total milli mole of OH- = 1+2 = 3
Total volume of solution(ml)=20+20=40ml
OH- CONCENTRATION =Milli mole of OH-/Total volume of solution(ml)=3/40M
pOH=-Log[OH-] = -log3/40 = log40-log3 =1.6020 -.4771 = 1.1249
pH=14–1.1249=12.8751
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