How can a substance with pH of 5 changed to pH of 7 by using 45% KOH?

a solution with pH =5 has

[H+] = 1E-5



1 liter has 1E-5 mol H+



to reach pH=7 you need 1E-5 mol KOH



1E-5 mol = 1E-5*(39+1+16) = 5.60E-04 KOH



so you need for each liter of original solution

5.6E-4/0.45 = 1.24E-03 g of KOH 45%How can a substance with pH of 5 changed to pH of 7 by using 45% KOH?
To me the question is rather why do you need so much KOH to change the pH by only 2 units... the answer to that is that the substance in question is a buffer that resist change in pH. Buffer solutions consist of a weak acid and its conjugate base. The resistive action is the result of the equilibrium between the weak acid and its conjugate base.