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

20 drops hope this

i think   the answer is c hope i'm right