Answer to Question #248653 in Inorganic Chemistry for Jellyace

Question

Resolve the concentration of 10 mL of 0.1 N hydrochloric acid accidentally added with 90 mL of distilled water. Answer with solution please. Thank you.

Answer 1

Solution:

C₁ = 0.1 N

V₁ = 10 mL

C₂ = unknown

V₂ = V₁ + V(H₂O) = 10 mL + 90 mL = 100 mL

Hence,

(0.1 N) × (10 mL) = C₂ × (100 mL)

C₂ = (0.1 N) × (10 mL) / (100 mL) = 0.01 N

C₂ = 0.01 N

Answer: The concentration of hydrochloric acid will be 0.01 N