A .288 g sample of an unknown monoprotic organic acid is dissolved in water and titrated with a .115 M sodium hydroxide solution. After the addition of 17.54 mL of base, a pH of 4.92 is recorded. The equivalence point is reached when a total of 33.83 mL of NaOH is added. What is the molar mass of the organic acid?

To find the molar mass of the organic acid, we can use the information provided in the question.

First, let's calculate the moles of NaOH that were added to reach the equivalence point. We know the molarity of NaOH solution and the volume added:

Moles of NaOH = Molarity * Volume
= 0.115 M * 33.83 mL

Next, let's calculate the moles of NaOH that were added to reach the pH of 4.92. We know the volume added at this pH:

Moles of NaOH at pH 4.92 = Molarity * Volume
= 0.115 M * 17.54 mL

Since NaOH and the organic acid react in a 1:1 ratio, the moles of NaOH added to reach the equivalence point will be equal to the moles of the acid in the sample.

Now, let's calculate the moles of the acid:

Moles of acid = Moles of NaOH at pH 4.92
= 0.115 M * 17.54 mL

To determine the molar mass of the acid, we need to divide the mass of the acid sample by the moles of the acid:

Molar mass = Mass of acid sample / Moles of acid

The mass of the acid sample is given as 0.288 g. We can convert it to moles using the molar mass of the acid (which we are trying to find) and the moles of acid:

0.288 g / (Molar mass of the acid) = Moles of acid

Now we can substitute the values and solve for the molar mass of the acid:

0.288 g / (Molar mass of the acid) = 0.115 M * 17.54 mL

Solving this equation will give us the molar mass of the unknown organic acid.

A generic equation can be written as follows:

RCOOH + NaOH ==> H2O + RCOONa
mols NaOH = L x M
mols RCOOH = mols NaOH (since the acid is monoprotic).
mols RCOOH = grams/molar mass.
You have mols and grams. Solve for molar mass.
The pH of 4.92 at 17.54 mL NaOH is extraneous information unless you are asked to calculate Ka for the acid.

moles OH- = (liters)(molarity)

(0.115 mol NaOH/L)(0.03383L) = 3.89x10-3 mol OH- used to reach the end point

Moles acid = moles OH-
moles of acid neutralized = 3.89x10-3 mol

0.288 g acid/0.03383L = 8.513g/L
(8.513g/L)/(0.115 mol/L) = ________g/mole

Yeah asking for Ka was the second part of the question, but it wasn't one of the review questions