please tell me if i am correct
a buffer is made using HF and NaF-with a pH of 3.00. compute the ratio of the conjugate base F- to weak acid HF.
I previously found Ka of HF which is 7.2x10^-4 and the ph of HF is 3.15.
the concentration of NaF would be 1.00x10^-2 ????
and so the ratio would be 1.00 over 7.2 x 10^-2 ??
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
3.00 - 3.14 + log B/A
solve for the ratio of B/A. I get 0.72 but check my work.
To determine the ratio of the conjugate base F- to weak acid HF in the buffer solution, we need to apply the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, the weak acid is HF (hydrofluoric acid), and the conjugate base is F- (fluoride ion). The pKa value of HF can be determined by taking the negative logarithm of the Ka value:
pKa = -log(Ka) = -log(7.2x10^-4) = 3.14
Given that the pH of the buffer is 3.00, we can set up the Henderson-Hasselbalch equation as follows:
3.00 = 3.14 + log([F-]/[HF])
Now, let's solve for the ratio [F-]/[HF]:
3.00 - 3.14 = log([F-]/[HF])
-0.14 = log([F-]/[HF])
To obtain the ratio, we need to convert the logarithmic form to exponential form:
10^(-0.14) = [F-]/[HF]
Using a calculator, we find the value of 10^(-0.14) to be approximately 0.477.
Therefore, the ratio of [F-]/[HF] in the buffer solution is approximately 0.477.
Now, let's address your other statement. You mentioned that the concentration of NaF is 1.00x10^-2. However, the concentration of NaF does not affect the ratio of [F-]/[HF] in the buffer solution. The ratio depends solely on the concentrations of the weak acid and its conjugate base ([HF] and [F-], respectively).
In summary, the ratio of the conjugate base F- to weak acid HF in the buffer solution is approximately 0.477, and the concentration of NaF does not affect this ratio.