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!
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.
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?
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.
the pH is NOT 1.85
To determine the pH of an HF solution, you need to use the equilibrium expression for the dissociation of HF. HF is a weak acid, so it partially dissociates in water.
The balanced equation for the dissociation of HF is:
HF ⇌ H+ + F-
The equilibrium expression is:
Ka = [H+][F-]/[HF]
Given that Ka has not been provided, we need to find the value of Ka for HF. The pKa value for HF is commonly known as 3.17.
Now let's solve the problem using the given information:
a) .280 M HF solution:
To determine the pH of this solution, we need to find the concentration of H+. Since HF partially dissociates, let's assume x M of HF dissociates into H+ and F-.
The concentration of H+ at equilibrium is x M, and the concentration of F- is also x M. The initial concentration of HF is 0.280 M, so the concentration of HF at equilibrium is (0.280 - x) M.
Substituting these values into the equilibrium expression:
Ka = [x][x]/[0.280 - x]
Since HF is a weak acid and the dissociation is less than 5%, we can make the approximation that 'x' is significantly small compared to 0.280, so we can neglect 'x' in the denominator.
Ka = [x][x]/[0.280]
Ka = x^2/[0.280]
The pKa value of HF is 3.17. Take the negative logarithm of Ka to find the value of pKa:
pKa = -log(Ka)
3.17 = -log(x^2/[0.280])
Now solve for x to find the concentration of H+:
x^2/[0.280] = 10^-3.17
x^2 = [0.280] * 10^-3.17
x = sqrt([0.280] * 10^-3.17)
Now that you have calculated the value of x, you can use it to find the pH of the solution:
pH = -log([H+])
pH = -log(x)
Calculate the value of pH using the value of x you found. This should yield the correct answer for part a. Remember to round the answer to the appropriate number of decimal places.
Continue the same process for parts b and c using their respective concentrations. Make sure to use the correct initial concentration of HF and solve for the concentration of H+ using the equilibrium expression. Finally, calculate the pH by taking the negative logarithm of the concentration of H+.
By following this approach, you should be able to find the correct answers for parts b and c as well.