Question

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 192.mg of oxalic acid H2C2O4, a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250.mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 55.8mL of sodium hydroxide solution.
Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

The molarity of the NaOH solution is 0.076 M

Explanation:

Step 1: Data given

Mass of oxalic acid = 192 mg = 0.192 grams

volume = 250 mL = 0.250 L

Molar mass oxalic acid = 90.03 g/mol

Step 2: The balanced equation

H2C2O4 + 2NaOH → Na2C2O4 + H2O

Step 3: Calculate moles of oxalic acid

Moles oxalic acid = 0.192 grams / 90.03 g/mol

Moles oxalic acid = 0.00213 moles

Step 4: Calculate molarity of oxalic acid

Molarity = Moles / volume

Molarity = 0.00213 moles / 0.250 L

Molarity = 0.00852 M

Step 5: Calculate Molarity of NaOH

2 Ca*Va = Cb*Vb

with Ca = Molarity of oxalic acid = 0.00852 M

with Va = volume of oxalic acid = 0.250 L

with Cb = Molarity of NaOH = TO BE DETERMINED

with Vb = volume of 0.0558 L

Cb = (2*0.00852 * 0.25) / 0.0558

Cb = 0.076 M

The molarity of the NaOH solution is 0.076 M