Should we use the PH=-log equation to solve this?

Hi!

According to given values in question we have C1=1.0mol/L and V1=10mL=0.01L and V2=1.0L we need to find the value of C2 in order to find the pH of resulting solution.

thus, C1 V1 = C2 V2.

1.0 x 0.01 = C2 x 1.0

C2 = 0.01 mol/L=[H+]

now, using pH=-log [H+]=-log[10^-2]=2.

Hope it helps:)

5 Likes