Respuesta :
Answer:
The volume of the concentrated nitric acid taken: V₁ = 4.3 mL
Explanation:
Given: Final pH of nitric acid (HNO₃) solution= 1.30
Volume of HNO₃ solution: V₂ = 600.0 mL
Concentration of stock HNO₃ solution: M₁ = 7.0 M
Volume of the stock HNO₃ solution: V₁ = ?
To find out the concentration of HNO₃ solution (M₂), we use the following equation: pH = - ㏒ [H⁺]
⇒ 1.3 = - ㏒ [H⁺]
⇒ ㏒ [H⁺] = - 1.3
⇒ [H⁺] = antilog (- 1.3)
⇒ [H⁺] = antilog (- 1.3) = 0.050 M = M₂
Now, to calculate the volume of concentrated 7.0 M HNO₃ (V₁) that should be added to prepare 600.0 mL of 0.050 M HNO₃ solution, we use the dilution equation: M₁ × V₁ = M₂ × V₂
⇒ V₁ = M₂ × V₂ ÷ M₁
⇒ V₁ = (0.050 M) × (600.0 mL) ÷ (7.0 M)
⇒ V₁ = 4.3 mL (rounded to two significant digits)
Therefore, the volume of the concentrated nitric acid taken: V₁ = 4.3 mL
