Chemistry, 16.12.2019 10:31 BeautyxQueen

1. if it takes 67 ml of 0.15 m naoh to neutralize 134 ml of an hcl solution, what is the concentration of the hcl? (5 points)

2. if it takes 27.4 ml of 0.050 m h2so4 to neutralize 357 ml of koh solution, what is the concentration of the naoh solution? (5 points)

3. if it takes 55 ml of 0.5 m naoh solution to completely neutralize 130 ml of sulfuric acid solution (h2so4), what is the concentration of the h2so4 solution? (5 points)

4. explain the difference between an endpoint and equivalence point in a titration. (5 points)

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Answer from: estherklein15ov3vx9

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄: The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl?

As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.The normality of the NaOH and HCl = Their molarity.

∵ (NV)NaOH = (NV)HCl

∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?

As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.

∵ (NV)H₂SO₄ = (NV)KOH

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

∴ The concentration of KOH = 7.675 mN = 7.675 mM.

Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?

As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of NaOH = Molarity of NaOH = 0.5 N.

∵ (NV)H₂SO₄ = (NV)NaOH

∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.

Q₄: Explain the difference between an endpoint and equivalence point in a titration.

The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.

Answer from: Quest

Idon’t know sorry i don’t know the answer i’m s sorry

Answer from: Quest

Ireally don’t know this

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1. if it takes 67 ml of 0.15 m naoh to neutralize 134 ml of an hcl solution, what is the concentrati...

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