What volume of 0.5M NaOH must be added to 50mL of .0.1M benzoic acid to make 100.0 mL of 0.05 M benzoate buffer that is pH 4.5?
2 answers
When you post a problem like this in the future you should include the pKa; otherwise, the answer I get won't be the same as the answer you get since texts differ in tabular values for Ka, Ksp, etc.
Let HB = benzoic acid
I will use pKa 4.2 but you should use the value from your text/notes.
You have 50 mL x 0.1M HB = 5 millimols.
First calculate millmoles acid and base needed.
pH = pKa + log (base)/(acid)
4.5 = 4.2 + log b/a
0.3 = log b/a
b/a = 2 or
base = 2*acid. This is equation 1(two unknowns)
The second equation is
acid + base = 100*0.05 = 5 millimols.
Solve those two equation simultaneously for a and b. I get approximately 1.67 for a and 3.33 for b. Plug that into the following:
.........HB + NaOH ==> NaB + H2O
I........5.....0........0.....0
add............x...............
C.......-x....-x........x.....x
E.......5-x....0........x
So you want x to be NaB(benzoate) or base and above base = 3.33
Therefore, you want to add 3.33 millimoles NaOH. Convert to mL by
M = millimols/mL. YOu know mmols and M, solve for mL. I ran through this quickly and came up with approx 7 mL of the 0.5M NaOH but you should confirm that AND remember that's an estimate and not an exact number.
I always like to check these to make sure of no error; therefore, I recommend you make an ICE chart as above, add the value you get (my estimate of 7 mL) and calculate the values for acid and base, plug these into the HH equation and see if you end up with a pH of 4.5. Post your work if you get stuck.
I will use pKa 4.2 but you should use the value from your text/notes.
You have 50 mL x 0.1M HB = 5 millimols.
First calculate millmoles acid and base needed.
pH = pKa + log (base)/(acid)
4.5 = 4.2 + log b/a
0.3 = log b/a
b/a = 2 or
base = 2*acid. This is equation 1(two unknowns)
The second equation is
acid + base = 100*0.05 = 5 millimols.
Solve those two equation simultaneously for a and b. I get approximately 1.67 for a and 3.33 for b. Plug that into the following:
.........HB + NaOH ==> NaB + H2O
I........5.....0........0.....0
add............x...............
C.......-x....-x........x.....x
E.......5-x....0........x
So you want x to be NaB(benzoate) or base and above base = 3.33
Therefore, you want to add 3.33 millimoles NaOH. Convert to mL by
M = millimols/mL. YOu know mmols and M, solve for mL. I ran through this quickly and came up with approx 7 mL of the 0.5M NaOH but you should confirm that AND remember that's an estimate and not an exact number.
I always like to check these to make sure of no error; therefore, I recommend you make an ICE chart as above, add the value you get (my estimate of 7 mL) and calculate the values for acid and base, plug these into the HH equation and see if you end up with a pH of 4.5. Post your work if you get stuck.