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MUST ANWSER ALL 3 please provide anwser in bold and please have it be 100% right

ID: 486658 • Letter: M

Question

MUST ANWSER ALL 3 please provide anwser in bold and please have it be 100% right I'm using sapling and these are my last problems thank you so much !!
1) if the Kb of a weak base is 4.5 x 10^-6 what is the pH of a 0.21 M solution of this base ?
2)A certain weak base has a Kb of 8.90 x 10^-7. What concentration of this base will produce a pH of 10.27?
3)The Ka of a monoprotic weak acid is 3.98 x 10^-3 What is the percent ionization of a 0.170 M solution of this acid ? MUST ANWSER ALL 3 please provide anwser in bold and please have it be 100% right I'm using sapling and these are my last problems thank you so much !!
1) if the Kb of a weak base is 4.5 x 10^-6 what is the pH of a 0.21 M solution of this base ?
2)A certain weak base has a Kb of 8.90 x 10^-7. What concentration of this base will produce a pH of 10.27?
3)The Ka of a monoprotic weak acid is 3.98 x 10^-3 What is the percent ionization of a 0.170 M solution of this acid ? MUST ANWSER ALL 3 please provide anwser in bold and please have it be 100% right I'm using sapling and these are my last problems thank you so much !!

2)A certain weak base has a Kb of 8.90 x 10^-7. What concentration of this base will produce a pH of 10.27?
3)The Ka of a monoprotic weak acid is 3.98 x 10^-3 What is the percent ionization of a 0.170 M solution of this acid ?

Explanation / Answer

1]

if the Kb of a weak base is 4.5 x 10^-6 what is the pH of a 0.21 M solution of this base ?

HOH H+ + OH-
Kb = [H+][OH-] / [HOH]
since [H+] = [OH-] , then
Kb = [OH-][OH-] / [HOH]
[OH-] = sqrt ( Kb [HOH] )
[OH-] = sqrt ( 4.5 10^-6 * 0.21 ) = 0.97 10^-3

pOH = -log_10 (0.097* 10^-3) = 3.01
pH = 14 - 3.01 = 10.99

2]

A certain weak base has a Kb of 8.90 x 10^-7. What concentration of this base will produce a pH of 10.27?

Kb = 8.90x10^-7 = [B+][OH-] / [BOH]
for pH = 10.27, the pOH = 3.73 and [OH-] = 10^-3.73 = 1.87x10^-4M
8.90x10^-7 = [B+][OH-] / [BOH]

8.90x10^-7 = (1.87x10^-4)^2 /[BOH]
[BOH] = 0.393M

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