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
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.
3 answers
How did you find the first answer?
http://www.jiskha.com/display.cgi?id=1302151175
1 answer