He ph value of 0.20 m solution of ch3nh2 at 298 k is [given that its ionisation constant kb is 5.0 × 10–5]
Answers
Thus the value of pH of the solution is pH = 11.50
Explanation:
We are given that:
- Molarity of solution = 0.20
- Temperature = 298 K
- Ionization constant = 5.0 × 10–5
SOlution:
Initial molar concentration,
CH3NH2 + H2O = 0.2
CH3NH3^+ = 0
OH^- = 0
Equilibrium molar concentration:
CH3NH2 + H2O = 0.2 - α
CH3NH3^+ = α
OH^- = α
According to the law of chemical equiblirium:
Kb = [CH3 NH3+] [ OH-] / [ CH3NH2]
Kb = α x α / 0.2 - α
Kb = α^2 / 0.2
Now put the value in the fomula:
5 x 10^-5 = α^2 / 0.2
α = √5 x 10^-5 x 0.2
α = 0.00316
Concentration of [ OH-]
[ OH-] = 3.16 x 10^-3 M
Concentration of [ H2O-]
[ H2O-] = K2 / [ OH]-
Put the values in the formula:
[H2O-] = 10^-14 / 3.16 x 10^-3 M
[H2O-] = 3.16 x 10^-12 M
Calculate pH value:
pH = - log [ H3O]
pH = - log (3.16 x 10^12)
pH = 12 - log 3.16
pH = 12 - 0.499
pH = 11.50
Thus the value of pH of the solution is pH = 11.50
Answer :-
Thus the value of pH of the solution is pH = 11.50
Explanation:
We are given that:
Molarity of solution = 0.20
Temperature = 298 K
Ionization constant = 5.0 × 10–5
SOlution:
Initial molar concentration,
CH3NH2 + H2O = 0.2
CH3NH3^+ = 0
OH^- = 0
Equilibrium molar concentration:
CH3NH2 + H2O = 0.2 - α
CH3NH3^+ = α
OH^- = α
According to the law of chemical equiblirium:
Kb = [CH3 NH3+] [ OH-] / [ CH3NH2]
Kb = α x α / 0.2 - α
Kb = α^2 / 0.2
Now put the value in the fomula:
5 x 10^-5 = α^2 / 0.2
α = √5 x 10^-5 x 0.2
α = 0.00316
Concentration of [ OH-]
[ OH-] = 3.16 x 10^-3 M
Concentration of [ H2O-]
[ H2O-] = K2 / [ OH]-
Put the values in the formula:
[H2O-] = 10^-14 / 3.16 x 10^-3 M
[H2O-] = 3.16 x 10^-12 M
Calculate pH value:
pH = - log [ H3O]
pH = - log (3.16 x 10^12)
pH = 12 - log 3.16
pH = 12 - 0.499
pH = 11.50
Thus the value of pH of the solution is pH = 11.50