1
Question
When 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M CH3COOH solution, the pH will be:
(use :(pKa)acetic acid=5, log2=0.3, log3=0.5, log5=0.7)
When 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M CH3COOH solution, the pH will be:
(use :(pKa)acetic acid=5, log2=0.3, log3=0.5, log5=0.7)
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Solution
The correct option is D 8.85
NaOH+CH3COOH→CH3COONa+H2O
Millimoles of NaOH reacting = molarity×volume(inL)
= 0.1×50× =5 mmol
This is the same as the amount of the acid reacting - i.e. 5 mmol, all the reactants are used up in the reaction.
Concentration of the salt CH3COONa produced:
molestotal volume(mL)
= 5100
=0.05 M
For a salt of a weak acid and a strong base:
pH = 12[pKw+pKa+logC]
pKw=14, pKa=5 (given) and C = 0.05M
After putting up the values we get
pH = 8.85
NaOH+CH3COOH→CH3COONa+H2O
Millimoles of NaOH reacting = molarity×volume(inL)
= 0.1×50× =5 mmol
This is the same as the amount of the acid reacting - i.e. 5 mmol, all the reactants are used up in the reaction.
Concentration of the salt CH3COONa produced:
molestotal volume(mL)
= 5100
=0.05 M
For a salt of a weak acid and a strong base:
pH = 12[pKw+pKa+logC]
pKw=14, pKa=5 (given) and C = 0.05M
After putting up the values we get
pH = 8.85
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