Question

Can someone explain where I go wrong?When I calculate this I get 14.78.

PRACTICE PROBLEM 20-4: Calculate the pH of a solution that results when 20.0 mL of 6.0-M NaOH (aq) is diluted with water to a final volume of 75.0 mL. Answer: pH = 14.20

Since NaOH is a strong base shouldnt it dissociate completely so OH- is 6.0 M.

Then Kw = [H3O+][OH-] = 1x10^-14

So [H3O+] is (1x10^-14) / (6.0M) which is 1.66667 x 10^-15

Then -log(1.66667 x 10^-15) is 14.78 so I am confused how they get 14.2

Answer 1

As the NaOH solution is diluted with water......the molarity of the NaOH solution must be changed.So you have to first calculate the new molarity of NaOH solution after dilution and then do the calculation as usual......

PRACTICE PROBLEM 20 -4 we calculate the numben mafes in the finst Sample NaoH Jiven vume of NaoH 20 = 20 mL 1σσο Holanity of NaoH 6-0 M mol/L 6.0 No 0moles Na 04 given Sample 20 KX6-0 molK 0-12 mol 1ution with water e the solutin AJ ter final vume = 75 mL Thus, the final malanity Na 0H Solufion af ter ution O-12 mo = 6 mel/L 75 6

Thus is Strong elec the comcen tratin CH Na 0H troly te Would be LOH =16 M poH - log CoH -log 1.6 =-0.20 pH 4- (-0-20) 14 POH 14+0.20 14.20 Aus 1