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