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0.1 mole of CH3NH2 (Kb = 5 x10–4)is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution?
- a)8 x 10-2 M
- b)8 x 10-11 M
- c)1.6 x 10-11 M
- d)8 x 10-5 M
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
0.1 mole of CH3NH2 (Kb = 5 x10–4)is mixed with 0.08 mole of HCl ...
The correct option (b) 8 x 10-11M
Explanation:
CH3NH2 (base) on reaition with HCI (acid) to give a salt of weak base and strong acid as CH3NH^+3CI-
Explanation:
CH3NH2 (base) on reaition with HCI (acid) to give a salt of weak base and strong acid as CH3NH^+3CI-
Most Upvoted Answer
0.1 mole of CH3NH2 (Kb = 5 x10–4)is mixed with 0.08 mole of HCl ...
-4) is dissolved in 1.0 L of water. Calculate the hydroxide ion concentration (OH-) in the solution.
To find the hydroxide ion concentration, we need to determine the concentration of the CH3NH2 in the solution.
The Kb value for CH3NH2 is 5 x 10^-4, which means that in water, CH3NH2 will partially ionize according to the following equation:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Let x be the concentration of the CH3NH2 that ionizes. Therefore, the concentration of CH3NH3+ and OH- formed will also be x.
The initial concentration of CH3NH2 is 0.1 mole/L, so the concentration of CH3NH3+ and OH- formed will be x mole/L.
Using the Kb expression for CH3NH2, we can write:
Kb = [CH3NH3+][OH-]/[CH3NH2]
5 x 10^-4 = x * x /0.1
Rearranging the equation:
x^2 = 5 x 10^-4 * 0.1
x^2 = 5 x 10^-5
x = sqrt(5 x 10^-5)
x = 7.07 x 10^-3
Therefore, the concentration of OH- in the solution is 7.07 x 10^-3 M.
To find the hydroxide ion concentration, we need to determine the concentration of the CH3NH2 in the solution.
The Kb value for CH3NH2 is 5 x 10^-4, which means that in water, CH3NH2 will partially ionize according to the following equation:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
Let x be the concentration of the CH3NH2 that ionizes. Therefore, the concentration of CH3NH3+ and OH- formed will also be x.
The initial concentration of CH3NH2 is 0.1 mole/L, so the concentration of CH3NH3+ and OH- formed will be x mole/L.
Using the Kb expression for CH3NH2, we can write:
Kb = [CH3NH3+][OH-]/[CH3NH2]
5 x 10^-4 = x * x /0.1
Rearranging the equation:
x^2 = 5 x 10^-4 * 0.1
x^2 = 5 x 10^-5
x = sqrt(5 x 10^-5)
x = 7.07 x 10^-3
Therefore, the concentration of OH- in the solution is 7.07 x 10^-3 M.
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0.1 mole of CH3NH2 (Kb = 5 x10–4)is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution?a)8 x 10-2 Mb)8 x 10-11 Mc)1.6 x 10-11 Md)8 x 10-5 MCorrect answer is option 'B'. Can you explain this answer?
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