Calculating Volume of Chemical equations

Question

A chemistry student weighs out 0.0939g of phosphoric acid H3PO4, a triprotic acid, into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000M NaOH solution.

Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

Answer 1

In order to calculate the volume (V) for NaOH solution you first need to calculate the # of mmol of NaOH that is needed to completely neutralize H3PO4 (hence equivalence point), because we know the molar concentration of NaOH. Recall the following relationship -

#mmol of a solute = volume (in mL) of the solution x molar concentration (in M) of the solute

We are using mmol here since in the end you want to calculate V of NaOH in mL (but use mol if you want to calculate V in L). Note that 1mol = 1000mmol.

Note that for each mmol of H3PO4 it can provide a total of 3 mmol of proton (H⁺) - they can all react with OH^- from NaOH until they are all consumed (neutralized). So how many mmol of H⁺ can 0.0939g of H3PO4 provide?

#mmol of H3PO4 = 0.0939(g)/97.994(g/mol) = 0.0009582mol = 0.9582mmol

#mmol of H⁺ = 3 x #mol of H3PO4 = 3 x 0.9582mmol = 2.8747mmol

So this 2.8747mmol of H⁺ is needed to react with OH^- (and thus NaOH) in order to reach the final equivalence point. Therefore the #mmol of OH^- and thus NaOH is also 2.8747mmol. Since we know the molar concentration of NaOH solution (0.2000M) thus V for NaOH solution is

V of NaOH solution (in mL) = #mmol of NaOH / molar concentration of NaOH = 2.8747 (mmol) / 0.2000 (M) = 14.3735mL.