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sweet-ann [11.9K]
3 years ago
5

3. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide:

Chemistry
1 answer:
charle [14.2K]3 years ago
6 0

Answer:

39.6 mL

Explanation:

Step 1: Write the balanced neutralization reaction

Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)

Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH

The molar mass of CH₃COOH is 60.05 g/mol.

2.78 g × 1 mol/60.05 g = 0.0463 mol

Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH

The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.

Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂

0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL

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A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t
ioda

Answer:

0.053 L  is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

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A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
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Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

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  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

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