Question

Consider a 0.238 M aqueous solution of sodium hydroxide, NaOH.

Answer 1

Answer:

The correct answer is a. 0.223 grams, b. 3.36 × 10²¹ and c. 2.79 × 10⁻³ mol.

Explanation:

a. The volume of the solution given is 23.46 ml or 23.46 × 10⁻³ L, the molarity of NaOH is 0.238 M or 0.238 mol/L. Now the number of moles of NaOH will be,  

Number of moles = Molarity × Volume in Liters of solution

= 0.238 mol/L × 23.46 × 10⁻³ L  

= 5.5835 × 10⁻³ mol

Now the mass of NaOH will be,

= Moles × Molecular mass of NaOH

= 5.5835 × 10⁻³ mol × 39.997 g/mol  

= 223.322 × 10⁻³ grams

b) The number of moles in 23.46 ml is 5.5835 × 10⁻³ mol

Now the number of hydroxide ions found will be,  

= No. of moles × NA  

= 5.5835 × 10⁻³ × 6.022 × 10²³

= 3.36 × 10²¹

c) The balanced chemical equation is,

2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂O

The number of moles of H₂SO₄ neutralized is,  

= Number of moles of NaOH taken/2

= 5.5835 × 10⁻³/2 mol

= 2.7917 × 10⁻³ mol