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How many milliliters of 0.100 M NaOH are required to neutralize the following solutions?

a.) 10.0 mL of 0.0500 M HCl
b.) 25.0 mL of 0.126 M HNO3
c.) 50.0 mL of 0.215 M H2SO4

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Answers:

  • a.) 10.0 mL of 0.0500 M HCl: 5.00 ml
  • b.) 25.0 mL of 0.126 M HNO₃: 31.5 ml
  • c.) 50.0 mL of 0.215 M H₂SO4: 215. ml

Explanation:

All the reactions are the neutralization of strong acids with the same strong base.

At the neutralization point you have:

  • number of equivalents of the base = number of equivalent of the acid

And the number of equivalents (#EQ) may be calculated using the normality (N) concentration and the volume (V)

  • # EQ = N × V

Then, at the neutralization point:

  • # EQ acid = N acid × V acid

  • # EQ base = N base × V base

  • # EQ acid = # EQ base

  • N acid × V acid = N base × V base

Also, you can use the formula that relates normality with molarity

  • N = M × number of hydrogen or hydroxide ions

a.) 10.0 mL of 0.0500 M HCl

  • The number of hydrogen ions for HCl is 1 and the number of hydroxide ions for NaOH is 1.

  • 10.0 ml × 0.0500 M × 1 = V base × 0.100 M × 1

         ⇒ V base = 10.0 ml ×0.0500 M / 0.100 M = 5.00 ml

b.) 25.0 mL of 0.126 M HNO₃

  • The number of hydrogen ions for HNO₃ is 1 and the number of hydroxide ions for NaOH  is 1.

  • 25.0 ml × 0.126 M × 1 = V base × 0.100 M × 1

         ⇒ V base = 25.0 ml ×0.126 M / 0.100 M = 31.5 ml

c.) 50.0 mL of 0.215 M H₂SO4

  • The number of hydrogen ions for H₂SO4 is 2 and the number of hydroxide ions for NaOH  is 1.

  • 50.0 ml × 0.215 M × 2 = V base × 0.100 M × 1

         ⇒ V base = 50.0 ml ×0.215 M × 2 / 0.100 M = 215. ml

5ml, 31.5ml & 215ml is the required volume of NaOH required to neutralize the given solutions.

How we calculate the required volume from molarity?

We will calculate the required volume to neutralize the acids by using the following formula:

M₁V₁x = M₂V₂y, where

M₁ = molarity of given base NaOH = 0.10M

V₁ = volume of given base NaOH = to find?

x = no. of hydroxide ions = 1

M₂ = molarity of acid

V₂ = volume of acid

y = no. of hydrogens

(1). To neutralize 10.0 mL of 0.0500 M HCl, we do calculation as:

0.10 × V₁ = 0.05 × 10 × 1

V₁ = 0.5/0.10

V₁ = 5 ml

(2). To neutralize 25.0 mL of 0.126 M HNO₃, we do calculation as:

0.10 × V₁ = 0.126 × 25 × 1

V₁ = 0.126×25/0.10

V₁ = 31.5 ml

(3). To neutralize 50.0 mL of 0.215 M H₂SO₄, we do calculation as:

0.10 × V₁ = 0.215 × 50 × 2

V₁ = 21.5/0.10

V₁ = 215 ml

Hence, 5ml, 31.5ml & 215ml is the required volume of NaOH.

To know more about molarity, visit the below link:

https://brainly.com/question/10608366

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