We know that, pH=log[H+]1 Hence, (pH)1=−log[H+]1 ?(i) (pH)2=−log[H+]2 ?(ii) Subtract eq. (i) from equation (ii): (pH)2−−(pH)1=−log[H+]2+log[H+]1 or (pH)+1−−(pH)1=log[H+]2[H+]1 or 1=log[H+]1/[H+]2 or [H+]2[H+]1=10(∵1=log10) Hence, [H+]2=101[H+]1 i.e. [H+] must be decreased 101th of its original volume.