Question

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.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 his sodium hydroxide solution. When the titration reaches the equivalence point, the student finds he has used 28.4mL 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 student's sodium hydroxide solution is 0.0219 M

Explanation:

1) Chemical reaction.

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

• sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

• NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

• 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

2) Mole ratio

• 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

3) Starting amount of oxalic acid

• mass = 28 mg = 0.028 g
• molar mass = 90.03 g/mol
• Convert mass in grams to number of moles, n:

        n = mass in grams / molar mass = 0.028 g / 90.03 g/mol =  0.000311 mol

4) Titration

• Volume of base: 28.4 mL = 0.0248 liter
• Concentration of base: x (unknwon)

• Number of moles of acid: 2.52 mol (calculated above)

• Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

$\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622$

That means that there are 0.000622 moles of NaOH (solute)

5) Molarity of NaOH solution

• M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using three signficant figures, such as the starting data are reported.

Answer 2

The molarity of the student's sodium hydroxide solution: 0.0219 M

Further explanation

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint / equivalence point of the reaction can use indicators according to the appropriate pH range

Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.

An acid-base titration there will be a change in the pH of the solution.

From this pH change a Titration Curve can be made which is a function of acid / base volume and pH of the solution

Acid-base titration formula

Ma Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

Sodium hydroxide and oxalic acid reaction

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

28.mg (0.028 g) of oxalic acid H₂C₂O₄ dissolves it in 250 mL (0.25 L) of distilled water

Then the mole H₂C₂O₄:

molar mass of H₂C₂O₄: 90

mol = mass: molar mass

mol = 0.028: 90

mol = 3.111.10⁻⁴

Molarity (M): mole: volume

M = 3.111.10-4: 0.25

M = 1,244.10⁻³

we suppose that a = NaOH, b = H₂C₂O₄, then

Ma = asked

Va = 28.4mL

na = 1 (NaOH produces 1 OH ion when dissociated -> NaOH -> Na⁺+ OH⁻)

Mb = 1,244.10⁻³

Vb = 250 ml

nb = 2 (H₂C₂O₄ produces 2 H⁺ions when dissociated ---> H₂C₂O₄ ---> 2H⁺ + C₂O₄²⁻)

Then :

Ma Va. na = Mb. Vb. nb

Ma 28.4. 1 = 1,244.10⁻³. 250 . 2

Ma = 0.0219

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