Question

How much liquid is needed to prepare 629.1mL of a solution that has a new concentration of 11.2M if the stock solution is 26.1M

Answer 1

270. mL

General Formulas and Concepts:

Acid-Base Titrations

Dilution: M₁V₁ = M₂V₂

• M₁ is stock molarity
• V₁ is stock volume
• M₂ is new molarity
• V₂ is new volume

Step-by-step explanation:

Step 1: Define

Identify

[Given] V₂ = 629.1 mL

[Given] M₂ = 11.2 M

[Given] M₁ = 26.1 M

[Solve] V₁

Step 2: Find Stock Volume

1. Substitute in variables [Dilution]: (26.1 M)V₁ = (11.2 M)(629.1 mL)
2. Multiply: (26.1 M)V₁ = 7045.92 M · mL
3. Isolate V₁ [Cancel out units]: V₁ = 269.959 mL

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs as our lowest.

269.959 mL ≈ 270. mL