12.85
1.70
[H+] = Kw / [OH-] = 1.0*10-14 / 2.5*10-4 = 4.0*10-11 mol/L
-log[1 X 10^-4 M OH(-)] = 4 14 - 4 = 10 pH ----------------
solution with [OH-] = 2.5 x 10-9 , A solution with [H+] = 1.2 x 10-4, A solution with pH = 4.5
HCN + H20 => CN- + H3O+ 0.45 X X Solve for X (4.0*10^-10) = x^2/0.45M x=1.34e-5= [OH] Find pOH= -log of the above number then 14-4.87=pH= 9.13
[OH-] = 3.31 log[OH-] = pOH = .51982 14-pOH = pH = 13.48
pH + pOH = 14. So pOH = 14 - 10.95 = 3.05 pOH = -log[OH-] [OH-] = 8.91 x 10-4 M
The pOH is 6,4.
An acidic solution will have a pOH ranging from 7 to 14
pOH = -log[OH-] = -log(5*10^-3) = 2.3 pH = 14 - pOH = 14.0 - 2.3 = 11.7 (at 25 oC)
pH and pOH are a measure of the concentration of the hydronium ions and hydroxyl ions respectively in the solution. pH = -log[H+] pOH = -log[OH-] and they are related: pH + pOH = 14
pH + pOH = 14. So pOH = 14 - 1.12 = 12.88 pOH = -log[OH-] [OH-] = 1.31 x 10-13 M
1.7
pOH= -log(0.0220M OH) pH=14-pOH pH=12.34242268
pH + pOH = 14pH = -log 3.60x10^-10 = 9.44pOH = 14 - 9.44 = 4.56
pH + pOH = 14. So pOH = 14 - 6 = 8 pOH = -log[OH-] [OH-] = 10-8 M
1.70