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?

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 KOH solution?

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?

The answer to your question is 1) 0.0075  2) 0.0077   3) 0.0105

Explanation:

1.-

Chemical reaction

               NaOH + HCl  ⇒    NaCl   +   H₂O

Calculate the moles of reactants

Molarity = moles / volume

moles = molarity x volume

moles = 0.15 x 0.067

          = 0.001005 moles of NaOH

               1 moles of NaOH 1 moles of HCl

               0.001005 moles    x

                x = (0.001005 x 1)/ 1 = 0.001005 moles of HCl

Concentration of HCl = 0.001005/0.134

                                   = 0.0075 M

2)

Chemical reaction

         H₂SO₄   +   2KOH  ⇒   K₂SO₄  +  2H₂O

Calculate the moles of H₂SO₄

moles = molarity x volume

moles = 0.05 x 0.0274

moles = 0.00137

        1 mol of H₂SO₄ 2 moles of KOH

        0.00137              x

         x = (0.00137 x 2) / 1

         x = 0.00274 moles

Calculate the molarity

Molarity = 0.00274/0.357

Molarity = 0.0077

3)

Chemical reaction

      2NaOH  +   H₂SO₄   ⇒    Na₂SO₄   +   2H₂O

Calculate the moles of NaOH

moles = molarity x volume

moles = 0.5 x 0.055

          = 0.0275

         2 moles of NaOH  1 mol of H₂SO₄

          0.0275 moles        x

           x = (0.0275 x 1)/2

           x = 0.01375 moles of H₂SO₄

Molarity = 0.01375/0.130

Molarity = 0.0105

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.