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Effectus [21]
2 years ago
13

In the titration of 82.0 mL of 0.400 M HCOOH with 0.150 M LiOH, how many mL of LiOH are required to reach the equivalence point

Chemistry
1 answer:
Igoryamba2 years ago
8 0

Answer:

218.7 mL

Explanation:

The reaction that takes place is:

  • HCOOH + LiOH → LiCOOH + H₂O

First we <u>calculate how many HCOOH moles reacted</u>, using the <em>given volume and concentration</em>:

  • 82.0 mL * 0.400 M = 32.8 mmol HCOOH

As <em>1 HCOOH mol reacts with 1 LiOH mol</em>, 32.8 mmoles of LiOH are needed to react with 32.8 mmoles of HCOOH.

Finally we <u>calculate how many mL of a 0.150 M solution would contain 32.8 mmoles</u>:

  • 32.8 mmol / 0.150 M = 218.7 mL
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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate
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0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: m_{(AcO)_2Cu} = 0.972 g

Volume of the sodium chromate solution: V_{Na_2CrO_4} = 150.0 mL

Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol

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n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

n_{(AcO)_2Cu} = 0.0053515 mol

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M

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