1. Calculate the molarity of a barium hydroxide solution if you used 41.65 mL of it to neutralize 1.190 g of potassium hydrogen phthalate. Ba(OH) 2 (aq) 2 KHC 8 H 4 O 4 (aq) BaC 8 H 4 O 4 (aq) K 2 C 8 H 4 O 4 (aq) 2 H 2 O(l) 2. Write a balanced chemical equation for the reaction of hydrochloric acid with sodium hydroxide. 3. Calculate the molarity of a hydrochloric acid solution if 34.21 mL of 0.0431M sodium hydroxide neutralizes 25.00 mL of the acid solution. 4. Write a balanced chemical equation for the reaction of phosphoric acid with sodium hydroxide. 5. Calculate the molarity of a phosphoric acid solution if 34.21 mL of 0.0431M sodium hydroxide neutralizes 25.00 mL of the acid solution.

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jufsanabriasa jufsanabriasa · Answer 1 · 2021-05-04T03:02:30+02:00

Answer:

1. 0.0700M barium hydroxide.

2. HCl + NaOH → H₂O + NaCl

3. 0.0590M HCl

4. H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

5. 0.01967M H₃PO₄

Explanation:

1. Based on the reaction:

Ba(OH)₂(aq) + 2KHC₈H₄O₄(aq) → BaC₈H₄O₄(aq) + K₂C₈H₄O₄(aq) + 2H₂O(l)

To find the molarity we have to convert the KHP to moles. With moles and volume we can find the molarity of the basrium hydroxide as follows:

Moles KHP -Molar mass: 204.22g/mol-

1.190g * (1mol / 204.22g) = 5.827x10⁻³ moles KHP

Moles Ba(OH)₂:

5.827x10⁻³ moles KHP * (1mol Ba(OH)₂ / 2mol KHP) = 2.91x10⁻³ mol Ba(OH)₂

Molarity:

2.91x10⁻³ mol Ba(OH)₂ / 0.04165L =

2. An acid reacts with a base to produce water and the salt. When HCl, reacts with NaOH, the reaction is:

3. The moles of NaOH added are:

34.21mL = 0.03421L * (0.0431mol / L) = 1.474x10⁻³ moles NaOH = Moles HCl

The molarity is:

1.474x10⁻³ moles HCl / 0.02500L =

4. The phosphoric acid, H₃PO₄, reacts with NaOH as follows:

5. The moles of NaOH added are:

34.21mL = 0.03421L * (0.0431mol / L) = 1.474x10⁻³ moles NaOH

The moles of phosphoric acid are:

1.474x10⁻³ moles NaOH * (1mol H₃PO₄ / 3mol NaOH) =

4.91x10⁻⁴ moles H₃PO₄. In 25.00mL = 0.02500L:

4.91x10⁻⁴ moles H₃PO₄ / 0.02500L =