Question

Show calculation for the preparation of 20 mL of reaction mix made of 0.8 g/L starch solution from 2 g/L starch stock solution and Phosphate/NaOH, pH 7 buffer. How much buffer should you add to get a final volume of 17.5 mL?

Answer 1

C₁ V₁ = C₂ V₂

C₁ = 2 g/L = 2 mg/mL   V₁ = ?   C₂ = 0.8 g/L = 0.8 mg/mL V₂ = 20 mL

2 * V₁ = 0.8 * 20

V₁ = 0.8 * 20 / 2

V₁ = 8 mL

Volume of buffer = Volume of reaction mix - Volume of stock solution

= 20 - 8

Volume of buffer = 12 mL

So, you have to take 8 mL of stock solution and add 12 mL of buffer to make 20 mL of 0.8 g/L reaction mix.

If the final volume of reaction mix is 17.5 mL, V₂ in the above formula takes the value of 17.5 mL. So,

C₁ = 2 g/L = 2 mg/mL   V₁ = ?   C₂ = 0.8 g/L = 0.8 mg/mL   V₂ = 17.5 mL

2 * V₁ = 0.8 * 17.5

V₁ = 0.8 * 17.5 / 2

V₁ = 7 mL

Volume of buffer = Volume of reaction mix - Volume of stock solution

= 17.5 - 7

Volume of buffer = 10.5 mL

So, you have to take 7 mL of stock solution and add 10.5 mL of buffer to make 17.5 mL of 0.8 g/L reaction mix.