Question

0.2 M HCl is used to neutralize 20 mL of 0.35 M NaOH.
How many moles of OH- are present?
How many moles of H+ are needed to neutralize the OH-?
What volume of 0.2 M HCl is needed?

Answer 1

Answer:

7×10⁻³ moles of OH⁻ ions are present in 20mL 0.35M NaOH.

7×10⁻³ moles of H⁺ are needed to neutralize that amount of OH⁻.

35mL of 0.2 M HCl is needed to neutralize  20mL 0.35M NaOH.

Explanation:

Given,

    volume of NaOH , V₂= 20mL = 0.02L

        molarity of NaOH = 0.35M

            molarity of HCl =0.2M

       number of moles = molarity × liters of solution

         moles of NaOH = 0.35 × 0.02 = 0.007 mol.

1 mole of NaOH contains 1 mole of OH⁻ ions. So 7×10⁻³ moles of OH⁻ ions are present.

1 mole of HCl is needed to neutralize 1 mole of NaOH.

So, 7×10⁻³ moles of H⁺ are needed to neutralize 7×10⁻³ moles of OH⁻.

Since the number of H⁺ in the HCl and number OH⁻ in the NaOH is one, we can write molarity = normality

for neutralization ,

               N₁V₁ = N₂V₂

           0.2× V₁ = 0.35N×20mL

⇒                  V₁ = 7 ÷ 0.2 = 35mL

35mL volume of 0.2 M HCl is needed