Posted by John on Wednesday, April 6, 2011 at 9:48pm.
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?
Related Questions
Chemistry - The Ka for hydroflouric acid is 4.5 *10^-4 at 298K. Calculate the ...
Chemistry - I can't figure out how to solve this problem: The Ka for ...
chemistry - 1)100ml sample of solution that is 0.2m in both Naf and Hf has 4.0 ...
chemsitry - 1)100ml sample of solution that is 0.2M in both Naf and Hf has 4.0 ...
Chemistry - Write a net ionic equation and calculate K for the reaction in aq ...
Chemistry - A buffer is prepared by mixing 45.0 mL of 0.15 M NaF to 35.0 mL of 0...
Chemistry - Calculate amount of solid NaF required to prepare a 250mL acidic ...
Chemistry - 44.70 ml of 0.100 M NaOH are required to completely neutralize 50.00...
chemistry - a student needed 1000 ml of a 0.10 buffer of ph 4.0. calculate the ...
Chemistry - Consider the titration of 100.0 mL of 0.260 M propanoic acid (Ka = 1...
For Further Reading