The Ka for hydroflouric acid is 4.5 *10^-4 at 298K. Calculate the amount of sold NaF that is required toprepare a 250. mL acidic buffer with pH = 2.75. The inital molarity of hydroflouric acid is .250M but you have only 75mL.
Part 2
Recalculate the pH of the buffer prepared in "a" above if 40.0 mL of 0.200 M NaOH is added. Print an equation that shows how the sodium hydroxide is consumed by the buffering agents; an appropriate arrow is expectd. Provide a statement why sodium ion is ignored. Show accountability ofthe conjugate weak acid and weak base. Use either the Ka or Henderson-Hasselbach equation.
NOTE: I Have already calculated the first part of this problem and calculated .2g of NaF are needed for that part.
OK. I worked the problem using your Ka value for HF and obtained 0.197 g which is close enough for me to think you probably worked it correctly. After the addition of 40 mL of 0.200M NaOH, the concns follow. The molarity of the HF is (75*0.25/290) = 0.06466M which you should confirm.
molarity of the NaF is (0.2/41.988/0.29) = 0.01643M which you should confirm.
Adding 40 mL of 0.2M NaOH = (40*0.2/290)= 0.02759 which you should confirm.
...............HF + OH^- ==> F^- + H2O
initial......0.06466 0......0.01643..0
add.................0.02759...........
change....0.02759...-0.2759..+0.02759..
equil........sum.....0........sum......
Plug these sum values into the HH equation and solve for pH.
Check all of these numbers. If I didn't goof the pH = 3.42
How did you find the first answer?
http://www.jiskha.com/display.cgi?id=1302151175
To solve part 2 of the problem, we need to consider the reaction between NaF and NaOH in the buffer solution. Here is the balanced equation:
NaF + NaOH => Na2O + HF
In this reaction, NaF reacts with NaOH to form Na2O and HF. The arrow indicates that the reaction goes from left to right.
Now, let's discuss why the sodium ion (Na+) is ignored. In a buffer solution, the purpose is to maintain a stable pH. This stability is achieved by the equilibrium between the weak acid (HF) and its conjugate weak base (F-). Adding NaOH will react with HF to form NaF, but the sodium ion does not participate in the buffering action. It simply serves as a spectator ion.
To calculate the pH of the buffer solution after addition of NaOH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this equation, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base (F-), and [HA] is the concentration of the weak acid (HF).
Given that the initial molarity of HF is 0.250 M and you have 75 mL of it, we can calculate the moles of HF:
moles of HF = initial molarity x volume in liters
moles of HF = 0.250 M x 0.075 L = 0.01875 moles
Since the NaF is the conjugate base of HF, the moles of NaF needed would be the same as the moles of HF. Thus, 0.01875 moles of NaF are required.
After adding 40.0 mL of 0.200 M NaOH, we can determine the concentration of NaF:
moles of NaF = initial moles of NaF + moles of NaOH
moles of NaF = 0.01875 moles + (0.200 M x 0.040 L) = 0.02575 moles
Next, we need to calculate the new volume of the buffer solution after the addition of NaOH:
volume of buffer = initial volume of HF + volume of NaOH
volume of buffer = 75 mL + 40 mL = 115 mL = 0.115 L
Now, we can calculate the concentration of NaF in the buffer solution:
concentration of NaF = moles of NaF / volume of buffer
concentration of NaF = 0.02575 moles / 0.115 L ≈ 0.224 M
Finally, we can substitute the values into the Henderson-Hasselbalch equation to calculate the new pH:
pH = pKa + log ([A-]/[HA])
pH = -log(4.5 * 10^-4) + log(0.224 / 0.250)
pH ≈ -3.35 + log(0.896)
pH ≈ -3.35 + (-0.048)
pH ≈ -3.35 - 0.048
pH ≈ -3.398
Therefore, the new pH of the buffer solution after the addition of 40.0 mL of 0.200 M NaOH is approximately 3.398.