a. Call phenol HP.
mols HP = grams/molar mass = approx 0.04 but you need a better number than that. Then M = mols/L = 0.04/0.065 = approx 0.6
.....HP ==> H^+ + P^-
I...0.6.....0......0
C....-x.....x......x
E...0.6-x...x......x
Substitute the E line into Ka for phenol and solve for x = (H^+), then convert to pH.
b.
HP + NaOH ==> NaP + H2O
You have mols HP initially.
Caculate mols NaOH added.
One will be in excess; probably HP but you need to confirm that.
Calculate how much of the excess reagent is present. That many mols/total volume in L will give you the M of the excess reagent. If HP is the excess then redo as in part a to calculate pH.
c. mL HP x M HP = mLNaOH x M NaOH.
Substitute and solve for mL NaOH
Phenol is a weak acid. We take 3.7 grams of phenol and dissolve it in 65 ml of water. To that water we add 40 ml of a 0.52 M solution of NaOH.
You will need to look up the Ka value for Phenol
a. What is the pH when the phenol is dissolved in the 65 ml of water ?
b. Once the base is added, what is the resulting pH ?
c. What is the total volume of base needed to reach the equivalence point ?
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