Find the molarity of KOH solution with a pH of 9.5?

Question

help please....explain very well

anything is welcomed

Answer 1

pH+pOH = 14

pOH = 14-pH

pOH = 14-9.5

pOH =4.5

[OH] = 10^-pOH

[OH] = 10^-4.5

[OH] = 3.1623*10^-5

KOH <=> K+  +  OH-

Because molar concentration OH- = molar concentration KOH,

Therefore molar conc KOH = 3.1623*10^-5M

Answer 2

This is the chemical equation that describes this solution

KOH --> K+ + OH

Recall that pH = -Log[H3O+]

[OH-][H3O+] = kw

kw is the autoprotolysis of water constant and its equal to 1 x 10^-14

since we know the pH we can solve for [H+] (an abbreviation for the hydronium ion (H3O+))

pH = -Log[H+]

9.5 = -Log[H+]

recall that n = 10^a means that logn = a

-9.5 = Log[H+]

10^-9.5 = [H+]

[H+] = 3.16227766 x 10^-10 M

now use the autoprotolysis of water equation to find the concentration of  hydroxide ion (OH-):

[H+][OH-] = 1 x 10^-14

[OH-] = (1 x 10^-14)/[H+] = (1 x 10^-14)/(3.16227766 x 10^-10 M) = 3.1623 x 10^-5 M

Since KOH is a strong base the concentration of OH- is the same as the concentration of KOH, which is 3.1623 x 10^-5 M.