Asked by Charles
Can someone please help me answer these questions? Is additional information other than what is given needed to solve? Thanks in advance.
1) Calculate the PH of a buffer containing 0.100 M propanoic acid, HC3H5O2 and .100 M NaC3H502 after the following have been added. Ka for acetic acid is 1.3 x 10^-5.
a. 10.0 mL of 2.0 M HCl (assume no volume change occurs when acid is added)
b. 10.0 mL of 2.0 M NaOH (assume no volume change occurs when base is added)
2) A hydrofluoric acid buffer solution has the following concentrations: [HF] = .18M, [F-]= .51M. If 35mL of .15M NaOH is added to 354 mL of the buffer, what is the resultant pH?
1) Calculate the PH of a buffer containing 0.100 M propanoic acid, HC3H5O2 and .100 M NaC3H502 after the following have been added. Ka for acetic acid is 1.3 x 10^-5.
a. 10.0 mL of 2.0 M HCl (assume no volume change occurs when acid is added)
b. 10.0 mL of 2.0 M NaOH (assume no volume change occurs when base is added)
2) A hydrofluoric acid buffer solution has the following concentrations: [HF] = .18M, [F-]= .51M. If 35mL of .15M NaOH is added to 354 mL of the buffer, what is the resultant pH?
Answers
Answered by
DrBob222
#1 has no volume.
#2.
millimols HF = 354 mL x 0.18 = about 64
mmols F^- = 354 mL x 0.51 = about 180
added NaOH = 35 mL x 0.15 = about 5
.........HF + OH^- ==> F^- + H2O
I........64....0.......180
add............5..............
C.......-5....-5.........+5
E.......59.....0.......185
Substitute into HH equation and solve for pH. NOTE: My "abouts" are estimates. You need to redo each more accurately.
#2.
millimols HF = 354 mL x 0.18 = about 64
mmols F^- = 354 mL x 0.51 = about 180
added NaOH = 35 mL x 0.15 = about 5
.........HF + OH^- ==> F^- + H2O
I........64....0.......180
add............5..............
C.......-5....-5.........+5
E.......59.....0.......185
Substitute into HH equation and solve for pH. NOTE: My "abouts" are estimates. You need to redo each more accurately.
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