Question #046fa

#HCl# is a strong acid and it will completely dissociate in water. However water will also dissociate to give #H^+# and #OH^-# ions (auto ionization).

Therefore, from #HCl# the concentration of #H^+# is #[H^+]_(HCl)=10^(-8)M# and from #H_2O# the concentration of #H^+# is #[H^+]_(H_2O)=10^(-7)M#.

The pH is calculated from both concentrations #[H^+]_(Total)=[H^+]_(H_2O)+[H^+]_(HCl)=10^(-7)M+10^(-8)M~~10^(-7)M#

Therefore the #pH=-log[H^+]_(Total)=-log10^-7=7#

In this case, the concentration of #HCl# is too small to change the pH of water. The solution stays neutral.

The previous contributor did use the correct method to answer the question and in theory his answer is correct, but the answer makes no sense due to #HCl# being an acid so must have a pH lower than 7.
Hence, there must be a mistake in the actual question that was asked and this can in fact be proven if one uses the equilibrium constant for water #K_w=[H^+][OH^-]=10^(-14)#.

This would imply that #[OH^-]=10^(-14)/(1xx10^(-8)#

#=1xx10^(-6)M#

#>[H^+]#

This is impossible for an acid it cannot have this.