how to Calculating the ka value?

Question

The question is: A solution of HBr contains 2 g of HBr per litre and has a pH of 2.5. What is the Ka and the PoH?

My chem teacher never taught us this and the exam is in 2 day. How do u find the Ka???

the answer is 4.49x10 to the -9

dont even get the basics...really really dont get it, There are a few of us trying to figer out how this work are chem teacher sucks.

Answer 1

2 g of HBr per liter equals 2/80.9 moles per liter or 0.0247 Molar.

The pH was 2.5 so you know that -log [H+] is 2.5. You should be able to determine that [H+] = 0.00316 M

Ka = [H+] [H+] / [HBr]
[H+] = 0.00316
[HBr] = 0.0247 - 0.00316 = 0.0215
Ka = 4.64 x 10^-4

Answer 2

Get ready to be taught.
For the problems that you will see with a weak MONOPROTIC acid ( 1 H+ per molecule)
Ka = x^2 / M.
M is the initial conc of acid.
In this case, you find it by dividing 2 grams by the mole wt of HBr.

x is the [H+] in solution. You DO have the pH, so you can compute this as 10^-pH, which is about 3x10^-3.

So, substitute the numbers in and solve.

BTW, pOH has the same definition as does pH, but it for the OH- ion. From water equilibrium,
pH + pOH = 14.

Answer 3

The dissociation equation is:

HBr ---> H+ + Br-

There is a 1:1 molar ratio between H+ and Br-, so those two values will be the same when you plug them into the Ka expression.

The answers given are correct in their technique, I just wanted to add the little bit above.

Best wishes on your test.