Consider a 0.238 M aqueous solution of sodium hydroxide, NaOH. a. How many grams of NaOH are dissolved in 23.46 mL?
b. How many individual hydroxide ions (OH) are found in 23.46 mL?
c. How many moles of sulfuric acid, H2SO4, are neutralized by 23.46 mL of 0.238 M NaOH(aq)? [Hint: begin by writing a balanced equation for this neutralization reaction.]

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