[Paulo1913]

How do I solve this?

Calculate the pH of a 0.015 mol per L of cyclohexylamine, C6H11NH2

Ka= 2.17*10-11

If any assumption is made justify that they are valid

[UG]

You doing Level 3 this year Paulo1913? Or Scholarship?

[Paulo1913]

Scholarship

[UG]

Good Luck! You in year 12?

[sjb]

What's the expression for Ka?

[Borek]

While not very difficult, this question nicely checks several concepts at once. Hard to point to an easy answer.

http://www.chembuddy.com/?left=pH-calculation&right=toc

[darien2009]

Calculate the pH of a 0.015 mol per L of cyclohexylamine, C6H11NH2

Ka= 2.17*10-11

solution :

pH = -log [H+]
 
[H+]=quadratic  root of Ka[acid]
     
       =quadratic  root  of (2.17*10-11)(0.015)
        = quadratic  root   of 3.255*10-13
       = 5.71*10-7
pH= -log (5.71*10-7)
    = 6.244 

[Borek]

pH= -log (5.71*10-7) = 6.244 

No, you got it wrong in several places. Correct answer is 11.38.

[vhpk]

Since C₆H₁₁NH₂ is an amine, it should dissociate to hydroxyde ion not proton. Moreover, the product of concentration and Ka is approximate to Kw, the dissociation of water should be cared. Understanding that the product of concentration and charge of anion is equal to the one of cation, you will calculate the right answer.
Answer:  pH = 7.763

[jpg28]

I'm pretty sure that the answer is close to what Borek said. I also got 11.38 for my pH. Here's how I did it:

Since cyclohexylamine is a weak base, an aqueous solution of this would have a pH > 7. Since it's a weak base, we therefore should use its K_b value instead of its K_a value. I got K_b = 4.61 x 10^-4.

Since its K_b value is not very very small, it's safer to use the quadratic equation to solve for the [OH^-] concentration, which, when solved, is equal to 2.41 x 10^-3 M.

Solving for the pOH of the solution, it would be around 2.62, and since we want its pH value, we subtract it from 14.

Thus, pH = 11.38.

[Borek]

Since cyclohexylamine is a weak base, an aqueous solution of this would have a pH > 7. Since it's a weak base, we therefore should use its K_b value instead of its K_a value. I got K_b = 4.61 x 10^-4.

Since its K_b value is not very very small, it's safer to use the quadratic equation to solve for the [OH^-] concentration, which, when solved, is equal to 2.41 x 10^-3 M.

Solving for the pOH of the solution, it would be around 2.62, and since we want its pH value, we subtract it from 14.

Thus, pH = 11.38.

And this is finally the correct answer. Have a snack

[vhpk]

I'm so sorry for not noticing that the value given is the Ka value not Kb. Ans pH=11.38 is correct