The pH of 0.02M solution of a weak acid was measured at 7.6. What is the [OH-] in this
solution?
B) The Ka for benzoic acid is 6.4 x 10-5M. 150 ml of 0.1 M NaOH is added to 200 ml of 0.1 M
benzoic acid, and water is added to give a final volume of 1 L. What is the pH of the final solution?
A). pH = -log(H^+)
7.6 = -log(H^+). Solve for (H^+). You should get approximately 2.5E-8.
B)
150 mL x 0.1M NaOH = 15 millimiles.
200 mL x 0.1M benzoic acid = 20 mmols.
Let's call benzoic acid HB, then
............HB + NaOH ==> NaB + H2O
initial....20......0........0.....0
add................15..............
change.....-15...-15........15...15
equil.......5......0.........15
pH = pKa + log (base)/(acid)
pH = 4.19 + log (15/5); solve for pH.
Note and this is important:
I used millimoles (the equivalent of mol) INSTEAD of concentration. The HH equation calls for concentration. I used millimoles as a shortcut. The answer will be the same; however, some profs will count off if you use mols (or millimoles) instead of concentration. So you need to do one of three things if your prof is picky (as I am).
(base) = mmols/mL = 15/1000 = ?
(acid) = mmols/mL = 5/1000 = ?
#1.
pH = 4.19 + log(0.015/0.005) = ?
#2.
pH = 4.19 + log(15/1000/5/1000) =
= cancel the denominator since both are the same (by the way they will ALWAYS be the same) so
= 4.19 + log(15/5) = ?
or third,
pH = 4.19 + log(15/volume/(5/volume) =
cancel v and
pH = 4.19 + log (15/5) = ?
To find the [OH-] in a solution, we need to know the pH and then use the formula for the calculation. The pH is a measure of the hydrogen ion concentration [H+], and it relates to the [OH-] through the equation:
pH + pOH = 14
Since we are given the pH, which is 7.6 in this case, we can find the pOH:
pOH = 14 - pH
pOH = 14 - 7.6
pOH = 6.4
To find the [OH-], we use the equation:
[OH-] = 10^(-pOH)
Substituting the value we found for pOH:
[OH-] = 10^(-6.4)
[OH-] ≈ 3.98 x 10^(-7) M
Therefore, the [OH-] in the solution is approximately 3.98 x 10^(-7) M.
Moving on to the second question about the pH of the final solution when NaOH is added to benzoic acid.
First, let's find the moles of benzoic acid and NaOH used. From the given concentrations and volumes:
Moles of benzoic acid = 0.1 M x 0.2 L = 0.02 moles
Moles of NaOH = 0.1 M x 0.15 L = 0.015 moles
Next, we want to determine which chemical is in excess to calculate the remaining moles for the reaction. In this case, NaOH is in excess because there are more moles of it compared to benzoic acid.
Since NaOH is a strong base, it will neutralize benzoic acid in a 1:1 stoichiometric ratio. Therefore, the remaining moles of benzoic acid after the reaction is:
Remaining moles of benzoic acid = Initial moles - Moles of NaOH used
Remaining moles of benzoic acid = 0.02 - 0.015
Remaining moles of benzoic acid = 0.005 moles
Now, let's calculate the concentration of benzoic acid in the final solution after adding water:
Concentration of benzoic acid = Remaining moles / Final volume
Concentration of benzoic acid = 0.005 moles / 1 L
Concentration of benzoic acid = 0.005 M
Since benzoic acid is a weak acid, we need to calculate the Ka value to find the pH. The Ka value is given as 6.4 x 10^(-5) M.
Using the equation for the dissociation of benzoic acid:
Ka = [H+][C6H5COO-] / [C6H5COOH]
We assume that the concentration of [H+] is equal to [C6H5COO-], and using the equation for the Ka value, we can solve for [H+]:
6.4 x 10^(-5) = [H+][0.005] / [0.005]
[H+] = 6.4 x 10^(-5) M
Finally, we can find the pH using the equation:
pH = -log[H+]
pH = -log(6.4 x 10^(-5))
pH ≈ 4.19
Therefore, the pH of the final solution after adding 150 ml of 0.1 M NaOH to 200 ml of 0.1 M benzoic acid and diluting to a final volume of 1 L is approximately 4.19.