# chemistry

posted by .

I keep getting wrong answers for this question.
Determine the pH of an HF solution of each of the following concentrations.
a) .280 M
b) 5.3*10^-2 M
c) 2.50*10^-2 M

No Ka has been given. I haven't attempted b or c yet because I want to at least solve a first. So far I got a answer of 3.75 and 1.88 and both are incorrect. I think I know how to solve for this. Using Ka and approximations. I just don't understand where I am going wrong. Thanks in advance!

• chemistry -

These can't be solved with a Ka value. Why don't you post your work for a and let me look at it. I'm sure I'll see immediately what is going on.

• chemistry -

Oh okay I thought you used Ka to solve these. That's what I'm doing wrong. I tried looking up the Ka for HF and multiplying it to .280 then take the -log of that number. The reason why I had to different numbers is because I found two different Ka's. If I can't use Ka do I just take the -log of .280?

• chemistry -

First I made a typo with my response. You certainly DO need Ka and that's the only way you can do it. I just didn't type in with instead of without.
My text hqas 7.2E-4 for HF.
...........HF ==> H^+ + F^-
initial.0.280.....0.....0
change....-x......x.....x
equil....0.280-x...x....x

Ka = 7.2E-4 = (x)(x)/(0.280-x)
Solve for x and convert to pH.
Probably you can get by without a quadratic equation for the 0.280but it may take one for the others, especially the very weak one.
Using this value I get 1.85 for the pH of the 0.280 M soln. Your text may give a different value for Ka.

• chemistry -

the pH is NOT 1.85

## Similar Questions

1. ### Algebra

- ok... we keep getting this one wrong: Solve the following equation for x. Write your answer as a fraction in simplest form. -6(x-5)=-2x-8-6(3x-5) What is it you don't understand about this?
2. ### Chemistry

What is the pH (to three figures) of a 0.1 M solution of sodium acetate given that the acid dissociation constant (Ka) of acetic acid is 1.8E-5?
3. ### Algebra

i have this question i have been working on two days no and cannot get the correct solution. I am to write and equation of the line containin the given point and parallel to the given line. I put my answer in the form of y=mx+b and …
4. ### chemistry

I have been at this question for at least three hours now. Consider the chemical system below. 2 NOCl(g) 2 NO(g) + Cl2(g) K = 1.6 10-5 What is the equilibrium concentration of nitrogen monoxide, [NO], given each of these initial conditions?
5. ### chemistry

I have this question: A 71.7-mL solution of 0.182 M NaOH is titrated with 0.2086 M HC. And I got the rest of the question of right...but for this one part. 4.) after addition of 50.0 mL of HCl i keep on getting 12.95 but I'm wrong...can …
6. ### Chemistry

Calculate the concentrations of all species in a 0.810 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4 x 10^-2 and Ka2 = 6.3 x 10^-8 I calculated SO32- = 0.8096 M, HSO3- = 3.58E-4, H2SO3 …
7. ### Chemistry

Glutamic acid is triprotic amino acid with pKa values of 2.23, 4.42, and 9.95. You are given a 0.100 M solution of it in its fully protonated form (H3A). How many moles of KOH must be added to 100 mL of this solution to make a solution …
8. ### Chemistry

Calculate the molarity of a solution prepared by dissolving 16 g of urea, NH2CONH2, in 39 g of H2O. The density of the solution is 1.3 g/mL. I've tried this question so many times but I keep getting it wrong please help me
9. ### Chemistry

Help me set up this problem. I have been looking at a few examples of similar problems and I keep getting the wrong answers. I'm getting really confused. Q: The standard free energy change ΔG∘ and the equilibrium constant …
10. ### Chemistry

A 10.0-mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution: a) 0.00 mL b) 10.0 mL c) 20.0 mL c) 30.0 mL d) 40.0 mL **I know how to solve the problem …

More Similar Questions