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2 years ago

Answer anybody, confused

Chemistry

1 answer:

jeka57 [31]2 years ago

7 0

I'm just taking a self educated guess of A sorry if I'm wrong :C

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2 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2

Nezavi [6.7K]

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.

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