A chemistry student weighs out 0.126g of lactic acid HC3H5O3 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0500M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.

28 mL of NaOH

Explanation:

First, we need to write the overall acid base reaction:

HC₃H₅O₃ + NaOH <> NaC₃H₅O₃ + H₂O

According to the above reaction, we can see that this reaction is already balanced and that the mole ratio between the acid and the base is 1:1. Therefore we can say the following:

n₁ = n₂  (1)

1 would be the acid, and 2 the base.

Now, in an acid base titration, when the titration reach the equivalence point, in this point, moles of each reactant is the same, and we use expression (1), but we can also use concentration and volume of both reactants. so, if we want to know the volume of the base, we need to rearrange the  above expression into an expression that use concentration and volume, which is the following:

n = M * V  (2)

If we replace this expression into (1) we have:

M₁V₁ = M₂V₂  (3)

From there, we just solve for the volume.

V₂ = M₁V₁ / M₂    or  V₂ = n₁ / M₂  (4)

And this is how we can calculate the volume of the base. Now, we need to know the moles used of the acid. For that, we need the molecular weight of the lactic acid which is 90.08 g/mol, so the moles are:

n₁ = 0.126 / 90.08 = 1.4x10⁻³  moles

Now, we have these moles, we can use (4) to get the volume of the base:

V₂ = 1.4x10⁻³ / 0.05

the answer is c i believe