calculate pH of solution containing 125mL of .606M NaOH in 15L of water?

Question

So in a question like this, you would always find the moles of acid/base first...then find moles of conj acid/base of the inital volume plus added volume of water?

Answer 1

the moles of NaOH = the moles of OH - = 0.125 * 0.606 = 0.07575
the moles of OH - in the new solution = 0.07575/(15 + 0.125 ) = 5*10^(-3) M
pOH = - log ( 5*10^ -3) = 2.3
=> pH = 14 - pOH = 11.7

Answer 2

ans is ( a) pH= 10 rationalization------------------------ given---- pH = 11 so [H+] = 10^-11/litre even as 9 volume parts of water is further then entire volume = 10 litres so [H+] = 10^11'10 litre = 10^-10 so pH = 10 answer