A student has a can of pop (acid) of unknown concentration. Upon examination it was found that a 10mL sample of the pop could be neutralized by 37mL of 1.3 x 10-6 M solution of base

Question

a) draw and fully label your set up and steps for the above titration

b) complete the calculation for the concentration of the acid (pop)

c) what is the pH of the acid

d) what is the pH of the base

Answer 1

a) I can't draw anything on this platform, so for this, you're on your own.

b) For the concentration of the acid, we would have to know if the acid is monoprotic, diprotic, triprotic, or polyprotic. We'll assume it is monoprotic and represent the acid as HA. We'll also assume the base is a typical base that has 1 OH^- such as NaOH, as opposed to Ca(OH)₂, or Al(OH)₃. The question is rather poorly written.

HA + OH- ==> H2O + A-

mols OH- used in titration = 37 ml x 1 L / 1000 ml x 1.3x10^-6 mol/L = 4.81x10^-8 mols OH-

mols HA present = 4.81x10^-8 mols

Concentration of acid = mols acid / liters of solution; 10 ml solution = 0.01 liters, thus 4.81x10^-8 mols / 0.01 L

= 4.81x10^-6 mol/L = 4.81x10^-6 M = concentration of acid

c) pH of acid = -log [H⁺] = -log 4.81x10^-6 assuming the acid is a strong acid and not a weak acid (again, this is not stated in the question. Again, this is a poorly written question)

pH = -log 4.81x10-6 = 5.32

d) pH of base (assuming it is something like NaOH, a strong base) would be...

[OH-] = 1.3x10^-6 M

pOH = -log 1.3x10^-6 = 5.89

pH = 14 - 5.89

pH = 8.11