Jump to content

Finding Ka


ecoli

Recommended Posts

I'm studying for this chem. quiz, and I'm having trouble with this concept the proffesor barely covered, yet is going to be on the quiz anyway.

 

Calculate the pH of the following solutions.

 

a.) 0.10 M CH3NH3Cl

 

b) 0.050 M NaCN

 

There are other examples, but I can't figure out how to calculate the Ka of the solutions.

Link to comment
Share on other sites

Well, let's see. Apply a few rules of stoichiometry here:

Let's use 0.125M benzoic acid (C6H5COOH). Given that its pH is 2.55 at equilibrium. The ionization equation would be like this: C6H5COOH => H3O+ + C6H5COO-.

antilog(2.55) gives us a [H3O+]=2.8*10^-3M. From the equation seen above we get 2.8*10^-3M of C6H5COO- for every 2.8*10^-3M of H3O+ produced. So you get that.

Now, because 2.8*10^-3M of H3O+ was produced that much C6H5COOH was ionized.

If 2.8 * 10^-3 M C6H5COOH was consumed then the final equilibrium concentration left would be the amount ionized subtracted from the initial concentration. We get 0.125M-0.0028M=0.1222M.

Go ahead and put all the numbers in your equation now.

Link to comment
Share on other sites

ok...that's what confused me. Apparently my textbook doesn't know that. Because they are some questions like that. Or maybe I'm misreading them.

 

Can you guys make sure I'm doing this right, then, please?

 

Question

 

Give a solution of an acid labeled 0.100 M HA and Ka = 1.8 x 10^-5 ?

 

[1.] Write the Equation representing the dissociation of HA in water.

[2.] Write the Equilibrium Expression for the reaction.

[3.] What is the K do we use? (Ka)

[4.] Fill in the chart: Initial Change @ Equilibrium

[H+]

[A-]

[HA]

[5.] Calculate the concentration of All the species in solution { [HA] [A-],[H+],[OH-] } @ equilibrium.

[6.] Calculate the pH and pOH of the solution.

 

Answers

 

1.) HA + H2O <=> H3O+ + A-

2) HA <=> H+ + A-

 

3) Ka = [HA+][A-] / [HA]

 

4) __I ________C_________E

[H+] 0_______+ x________ x

[A-] 0 _______+ x___ ____ x

[HA] .100M___ - x _____.100 - x

 

5) Ka = 1.8 X 10^-5 = x^2 / 100 M - x SO x = 1.3 X 10^-3

 

[H+] = 1.3 X 10^-3 M

[HA+] = .100 M

[A-] = 1.3 X 10^-3 M

[OH] = 7.9 X 10^-12

 

6) pH = -log(1.3 X 10^-3) = 2.88

 

pOH = 14.0 - 2.88 = 11.1

Link to comment
Share on other sites

ok...that's what confused me. Apparently my textbook doesn't know that. Because they are some questions like that. Or maybe I'm misreading them.

 

Can you guys make sure I'm doing this right' date=' then, please?

 

Question

 

Give a solution of an acid labeled 0.100 M HA and Ka = 1.8 x 10^-5 ?

 

[1.'] Write the Equation representing the dissociation of HA in water.

[2.] Write the Equilibrium Expression for the reaction.

[3.] What is the K do we use? (Ka)

[4.] Fill in the chart: Initial Change @ Equilibrium

[H+]

[A-]

[HA]

[5.] Calculate the concentration of All the species in solution { [HA] [A-],[H+],[OH-] } @ equilibrium.

[6.] Calculate the pH and pOH of the solution.

 

Answers

 

1.) HA + H2O <=> H3O+ + A-

2) HA <=> H+ + A-

 

3) Ka = [HA+][A-] / [HA]

 

4) I C E

[H+] 0 + x x

[A-] 0 + x x

[HA] .100 M - x .100 - x

 

5) Ka = 1.8 X 10^-5 = x^2 / 100 M - x SO x = 1.3 X 10^-3

 

[H+] = 1.3 X 10^-3 M

[HA+] = .100 M

[A-] = 1.3 X 10^-3 M

[OH] = 7.9 X 10^-12

 

6) pH = -log(1.3 X 10^-3) = 2.88

 

pOH = 14.0 - 2.88 = 11.1

 

Everything looks fine to me in questions 1-4. In question number five, in the equation you're solving for x which is the H+ concentration. The A- concentration is exactly the same, but you forgot to solve for the HA concentration. HA should be 0.100 - 0.0013 = 0.0987 M.

 

Otherwise it looks good to me.

Link to comment
Share on other sites

  • 3 weeks later...

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.