An aqueous solution of a strong monoprotic acid has an initial volume V and a pH of 3.0 . Water is added to this solution until its pH reaches 3.6 . The volume of water added is: (Given 2=0.30 )

Step-by-step solution

Initial pH = 3.0 [ H ⁺]_ i = 10⁻³ M Final pH = 3.6 [ H ⁺]_ f = 10^ -3.6 M The ratio of final to initial concentration is: [ H ⁺]_ f [ H ⁺]_ i = 10^ -3.6 10⁻³ = 10^ -0.6 Given 2 = 0.30 , we have 4 = 2 0.30 = 0.60 . Thus, 10^ 0.6 = 4 10^ -0.6 = 1 4 . So, [ H ⁺]_ f = [ H ⁺]_ i 4 . Using the dilution formula M₁V₁ = M₂V₂ : V₂ = M₁V₁ M₂ = [ H ⁺]_ i V [ H ⁺]_ i 4 = 4V The final volume is 4V . The volume of water added is: V_ added = V₂ - V₁ = 4V - V = 3V Answer: 3V