Citric acid reacts with NaOH or KOH. The % yield of this reaction is _____. edit

Question

Citric acid reacts with NaOH or KOH. The % yield of this reaction is _____.

edit:I havent found the molarity of KOH but here are my runs I did with some of my 0.1M KHP in 100ml and some of my 0.9M KOH of 150ml.

run1:It took 2.35ml of KOH to neutralize 8.5ml of KHP

run2:It took 2.05ml of KOH to neutralize 15.6ml of KHP

Using these neutralization runs,the average Molarity of KOH can be found.
That average molarity will be need to find the average molarity of citric acid for the next titration that was done.

The next titration was titrating citric acid with KOH to find the average molarity of citric acid.

here are my runs I did with some of the KOH solution from the previous solution along with a citric acid solution that I used with flat soda.

run1:It took 0.4ml of KOH to neutralize 8.8ml of citric acid

run2:It took 0.3ml of KOH to neutralize 6.38ml of citric acid

I hope this data and results will be enough information.

Explanation / Answer

Ans. Part 1: Standardization of KOH:

Trial 1: Given, [KHP] = 0.10 M

# Balanced reaction: KHC8H4O4(aq) + KOH(aq) -------> K2C8H4O4(aq) + H2O(l)

According to the stoichiometry of balanced reaction, 1 mol KOH neutralizes 1 mol KHP (KHC8H4O4).

That is, at titration end point-

            C1V1 (KHP) = C2V2 (NaOH)

            Or, 0.10 M x 8.5 mL = C2 x 2.35 mL

            Or, C2 = (0.10 M x 8.5 mL) / 2.35 mL

            Hence, C2 = 0.3617 M

Therefore, molarity of NaOH in trial 1 = 0.3617 M

# Trial 2- C1V1 (KHP) = C2V2 (NaOH)

            Or, C2 = (0.10 M x 15.6 mL) / 2.05 mL

            Hence, C2 = 0.7610 M

Therefore, molarity of NaOH in trial 2 = 0.7610 M

# Average molarity of NaOH = (0.3617 M + 0.7610 M) / 2 = 0.56135 M

Note: Trial 2 consumes less NaOH while volume of KHP is nearly doubled. It is experimental error – please recheck your experimental values and do calculations as presented.

# Part 2: Determination of [Citric acid]:

Balanced reaction: 3 NaOH + H3C6H5O7 --------> Na3C6H5O7 + 3 H2O

According to the stoichiometry of balanced reaction, 3 mol NaOH is required to neutralize 1 mol citric acid.

So, at the titration end point-

            C1V1 (KOH) = 3 x C2V2 (citric acid)           ; [3 denotes 1 citric acid gives 3 H+]

# Trial 1: C2 = (0.56135 M x 0.4 mL) / (8.8 mL x 3) = 0.00851 M

Therefore, [Citric acid] in trial 1 = 0.00851 M

# Trial 2: C2 = (0.56135 M x 0.3 mL) / (6.38 mL x 3) = 0.00880 M

Therefore, [Citric acid] in trial 2 = 0.00880 M

Now,

            Average molarity of citric acid = (0.00851 M + 0.00880 M) / 2 = 0.008655 M

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