[OH-] = 1x10^-3 M
[H+][OH-] = 1x10^-14
[H+] = 1x10^-14/1x10^-3 = 1x10^-11
pH = -log 1x10^-11 = 11
Done another way:
pOH = -log [OH-] = -log 1x10^-3 = 3
pH + pOH = 14
pH = 14 - 3 = 11
The pH of 1 M KOH is 13.0
The pH of 0,1 M KOH is 1,3
11 or 11.0 it doesnt matter both are right answers
14.
Calculus:
pOH = -log(conentration of OH-)= -log(1.0)= 0.0
pH = 14-pOH = 14 -0.0 = 14.0 (at 250C)
pOH= -log[0,01]= 2 pH= 14-pOH= 14-2 = 12
the pH value is 13.
11
11
9
-log(1.0 X 10^-4 M KOH) = 4 14 - 4 = 10 pH KOH ----------------
9
concentrated solution of NaOH and KOH have the pH value about 13.
0.04 M KOH produces an OH- concentration of 0.04 M. Thus, the pOH is -log 0.04 = 1.4 and the pH will be 14 - 1.4 = 12.6
9
-log(1.0 X 10^-4 M KOH) = 4 14 - 4 = 10 pH KOH ----------------
9
The pH of water increase.
concentrated solution of NaOH and KOH have the pH value about 13.
-log(3.5 X 10^-4 M) = 3.4559 14 - 3.4559 = 10.5 pH
0.04 M KOH produces an OH- concentration of 0.04 M. Thus, the pOH is -log 0.04 = 1.4 and the pH will be 14 - 1.4 = 12.6
Yes, a concentrated solution of NaOH and KOH may have pH=14
1/103 = 0.001 M ========( pH 3 ) 1/105 = 0.00001 M ============( pH 5 ) As you see, a pH of 3 has a 100 times concentration of 5 pH ( 10 * 10 devalued ) This is the scale; logarithmic.
You think probable to potassium chloride.
pOH = -log(10)[OH^-] pOH = -log(10) [ 48. x 10^-2] pOH = - (-1.31875...) pOH = 1.31875... pH = 14 - pOH pH = 14 - 1.31875... pH = 12.68124... pH = 12.68
[H3O+]= Kw/[OH-] = 1.00*10-14/6.70*10-2 =1.49*10-13 MpH= -log [H3O+] = -log (1.49*10-13)=12.8