A chemist must prepare 600.0 mL. of nitric acid solution with a pH of 1.30 at 25 He will do this in three steps: Fill a 600.0 mL. volumetric flask about halfway with distilled water Measure out a small volume of concentrated (7.0M) stock nitric acid solution and add it to the flask Fill the flask to the mark with distilled water. Calculate the volume of concentrated nitric acid that the chemist mugt measure out in the second step. Round your answer to 2 signifcant digits ml.

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