Calculate the [OH-] of each aqueous solution with the following [H30+]?

Question

Can i get the steps to solving?

Cleanser, 5.0 x 10 -10M

Answer 1

[H3O+][OH-] = 1.0 x 10^-14 M always at 25C
so if [H3O+]= 5.0 x 10^-10M
[OH-]= 2.0x10^-5

Answer 2

as long as you may submit this question, you have get right of entry to to laptop. in the accessory classes, you're able to have the potential to discover calculator.exe. starts calculator.exe, on the funcation bar, clicks on view and click on clinical (short cut back key, Alt-V, S and Alt-V, T), then you actual might have a clinical calculator, which supply "log", "Exp" and different progressed purposes. returned on your question, pH = - log ( [H3O+] ) , [H3O+] * [OH-] = 10^-14 , [OH-] = 10^-14 / [H3O+] a million. baking soda [H3O+] = a million.0*10^-8, [OH-] = 10^-14 / 10^-8 = 10^-6 ; pH = 8, pOH = 6 2. orange juice [H3O+] = 2.a million*10^-4, [OH-] = 10^-14 / 2.a million*10^-4 = (10/2.a million)*10^-15/10^-4 = 4.seventy six *10^-11; pH = 3.sixty 8, pOH = 10.32 3.milk [H3O+] = 4.9*10^-7, [OH-] = 10^-14 / 4.9*10^-7 = (10/4.9)*10^-15/10^-7 = 2.04 *10^-8; pH = 6.31, pOH = 7.sixty 9 4.bleach [H3O+] = 4.9*10^-12, [OH-] = 10^-14 / 4.9*10^-12 = (10/4.9)*10^-15/10^-12 = 2.04 *10^-3; pH = 11.31, pOH = 2.sixty 9

Answer 3

[OH-] = 2 x 10^-5