If 1.00L of muriatic acid w / a pH of 2.5 is poured to 8.00L water, what is the new molar concentration of the muriatic solution? Hint: use C1•V1 = C2•V2

The answer to your question is: C2 = 0.0004 M

Explanation:

Data

pH = 2.5; V = 1.0 L

V2 = 8.0 L C2 = ?

Formula

C1V1 = C2V2

C2 = C1V1 / V2

pH = - log[H⁺]

Process

[H⁺] = antilog - pH

[H⁺] = antilog (-2.5)

[H⁺] = 0.003 M = C1

Finally

(0.003) (1 l) = C2 (8)

C2 = 0.003 / 8

C2 = 0.0004 M