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.

No such thing as Ka of the solution. Ka describes dissociating acid. Effect of the acid presense in the solution is a pH change.

 

Take a look at these pH calculation lectures.

 

Best,

Borek

--

Chemical calculators at www.chembuddy.com

pH calculation

concentration conversion

don't you need the Ka , as per the equations

 

Ka = [H+][A-] / [HA]

 

pH = - log[H+]

 

Given the molarity, how do you find Ka?

can anyone help me out? My quiz is tommorow.

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.

Tell you what, here's a better way to remember the definition of Ka.

Ka = [H+][A-] / [HA]=(x)(x)/M-(x)

That will help you see the relationship between the values, as well as summarizing what I said above.

I understand both those equations, thank you for the explaination. I want to know, is it possible to find the Ka of an acid just given the molarity of a solution?

I understand both those equations, thank you for the explaination. I want to know, is it possible to find the Ka of an acid just given the molarity of a solution?

 

 

No. It is not possible.

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

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.

Ok, I see my mistake. I can't believe I forgot to do that. It's why I've had an uncomfortable feeling about this problem from the beginning. Thanks a lot for the help. I should be fine for my quiz tommorow.

No. It is not possible.

 

I finally figured out what they want. You are given the Ka of HCN, a strong base and fully dissociative, so now you can find the Kb, and from there solve for the pH.