Chemistry question?

Question

how would you calculate the pH of 0.5M NaOH, or 2 M HCl, or 3e-8 M HCl?

Answer 1

1)pH of 0.5M NaOH

NaOH---> Na+ + OH- strong acid so completely dissociated

the concentration of OH- is 0.5
so pOH = log 1/OH- = log 1/0.5 = log 2 = 0.3
And knowing that pH + pOH =14 pH = 14-0.3 = 13.7


2)pH of 2M HCl

HCl---> H+ + Cl- strong acid so concentration of H+ =2

pH = log 1/2 = -0.3 (But problem, true theoretically, but not in practice since you must know the activity of H+)

3) for HCl 3e-8 the concentration of H+ ions given by the acid is 3 10^-8
inwater you have 10^-7 ions H+
so the total amount of H+ is 1.3 10^-7

pH = log1/1.310^-7 = 6.89