Question

 

We titrate a juice ingridents are water, fructose, citric acid, and flavoring. I measure out 5mL of the juice and put it in a titrator machine. The known base is 0.1M NaOH, it takes 1.892mL to bring the pH up to 8.32.

I know the ratio of citric acid to sodium hydoxride is 3:1. I forget how to include this part in my equation... So this is what I do .1M*.001892L/.005L= .03784M of citric acid.  

Now I want to know the pH so I take pH=-log[.03784] and it is 1.42.  

I know this isn't the pH because we a pH meter that measures the pH of the juice and it is around 2.5-2.6 pH. Is this not possible or am I doing something wrong?

Answer 1

Ans. 3 mol NaOH is required to neutralize 1 mol citric acid.

So, at the titration end point-

            3 x (C1V1, NaOH) = (C2V2, citric acid)

            Or, 3 x 0.1 M x 1.892 mL = C2 x 5.0 mL

            Or, C2 = 0.113520 M

Therefore, [Citric acid] in juice = 0.113520 M

# Citric acid has three acidic H-atoms. The pKa values are 3.15, 4.77, and 5.19, respectively.

# For first Protonation: H₃C₆H₅O₇ ----------> H₂C₆H₅O₇^- + H⁺

pKa1 = 3.15

Ka1 = antilog (-pKa) = antilog (-3.15) = 7.0795 x 10^-4

Create an ICE table as shown in figure-

            Ka = [H₂C₆H₅O₇^-] [H⁺] / [H₃C₆H₅O₇]

            Or, 7.07946 x 10^-4 = (X) (X) / (0.113520 -X)

            Or, X² + (7.0795 x 10^-4)X + (8.0366 x 10^-5) = 0

Solving the quadratic equation we get following two roots-

            X1 = 0.00862            ; X2 = -0.00932

Since concertation can’t be negative, reject X2.

Thus, X = 0.00862

And, [H⁺] at equilibrium = X = 0.00862 M

Also, [H₂C₆H₅O₇^-] at equilibrium = X = 0.00862 M

# For second Protonation: H₂C₆H₅O₇^- ------> H₂C₆H₅O₇^2- + H⁺

Since pKa1 and pKa2 a re comparable, we also need to account the second protonation.

pKa2 = 4.77

Ka2 = antilog (-pKa) = antilog (-4.77) = 1.6982 x 10^-5

Create an ICE table as shown in figure-

            Ka = [H₂C₆H₅O₇^2-] [H⁺] / [H₃C₆H₅O₇^-]

            Or, 1.6982 x 10^-5 = (X) (X) / (0.00862 -X)

            Or, X² + (1.6982 x 10^-5)X + (1.4638 x 10^-7) = 0

Solving the quadratic equation we get following two roots-

            X1 = 0.0003742                    ; X2 = -0.0003911

Since concertation can’t be negative, reject X2.

Thus, X = 0.0003742

And, [H⁺] at equilibrium = X = 0.0003742 M

# Now,

Total [H⁺] after 2^nd protonation = [H⁺] after 1^st protonation + [H⁺] after 2^nd protonation

                                                            = 0.00862 M + 0.0003742 M

                                                            = 0.0089942 M

Now,

            pH = -log [H⁺] = -log (0.0089942) = 2.05

Note: Please check your values. At titration endpoint, the pH should have been near to 7.0. Your obtained pH of 8.32 is most likely to be possible the errors in theoretical and calculated pH of citric acid juice.