15.0 mL of .275 M sulfuric acid solution "completely neutralizes" 20.0 mL of sodium hydroxide solution. What is the concentration of the sodium hydroxide solution

The answer to your question is  0.415 M

Explanation:

Data

volume of H₂SO₄ = 15 ml

[H₂SO₄] = 0.275

volume of NaOH = 20 ml

[NaOH] = ?

Process

1.- Write the balanced chemical reaction

                    H₂SO₄  + 2NaOH   ⇒   Na₂SO₄  +  2H₂O

2.- Calculate the molar mass of H₂SO₄ and the molar mass of NaOH

molar mass of H₂SO₄ = 2 + 32 + (16 x 4) = 98g

molar mass of NaOH = 23 + 16 + 1 = 40 g

3.- Calculate the moles of H₂SO₄ in 15 ml

moles = Molarity x volume

moles = 0.275 x 0.015

moles = 0.0041

4.- Use proportions to calculate the moles of NaOH

                  1 mol of H₂SO₄ 2 moles of NaOH

                 0.0041 moles      x

                   x = (0.0041 x 2) / 1

                   x = 0.0083 moles of NaOH

5.- Calculate the molarity of NaOH

Molarity = moles/volume

Molarity = 0.0083/0.02

Molarity = 0.415

the increase of protons   hope this

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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