A solution of potassium chloride has a pH of 7.81.What is the [OH-] in mol/L ?
Can you please show steps I would appreciate it very much.
Thank you
pH = -log(H^+)
7.81 = -log*H^+)
log(H^+) = -7.81
Take the antilog of both sides.
(H^+) = 1.5488 x 10^-8 (rounded to 1.55*10^-8)
(H^+)(OH^-) = 1 x 10^-14
solve for (OH^-) = 6.46 x 10^-7
But there is an easier way to do it.
pH = 7.81
pH + pOH = 14; therefore,
pOH = 14 - 7.81 = 6.19
6.19 = -log(OH^-)
log(OH^-) = -6.19
(OH^-) = 6.46 x 10^-7
Check my work.