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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Alexander980 Alexander980
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9 years ago
Calculate [H+] and [OH-] for solutions with the following pH values
a) 4
b)8.52
c)12.60

A buffer is prepared in which the ratio [HCO3-] / [CO3^2-] is 4.0
a) What is the pH of this buffer (Ka HCO3- = 4.7 x 10^-11

b) What is the pH of the buffer if .0100 moles of HCL are added?

c) What is the pH of the buffer if .0200 moles of NaOH are added?
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wrote...
#1
9 years ago
Quote from: Alexander980 (9 years ago)
Calculate [H+] and [OH-] for solutions with the following pH values
a) 4
b)8.52
c)12.60

pH = - log ( [H+])

 a) - log ( [H+]) = 4

∴  [H+] = 10-4

  also [H+] [OH-]  = Kw = 10-14

=>  [OH-]  =10-10

 

b) - log ( [H+]) = 8.52

            ∴  [H+] = 3.019 x 10-9

             [OH-]   = 3.311 x 10-6

 

c)  - log ( [H+]) = 12.6

            ∴  [H+]  = 2.512 x10-13

                [OH-]   =3.98 x 10-2
wrote...
#2
9 years ago
If pH = 3.42
[H+] = 10^-3.42
[H+] = 3.80*10^-4M

To calculate [OH-]:
Equation:
[H+] [OH-] = 10^-14
[OH-] = 10^-14 /( 3.80*10^-4)
[OH-] = 2.63*10^-11

If pH = 0.00
[H+] = 10^0.00
[H+] = 1.00M

[OH-] = 10^-14/1.00
[OH-] = 10^-14M
wrote...
#3
9 years ago
The formula for pH is :
pH = (-log[H+])
and pOH = 14 - pH

i) 3.42 = (-log[H+])
=> [H+] = 10^-3.42

10.58 = (-log[OH-])
=> [OH-] = 10^-10.58


ii) 0 = log[H+]
=> [H+] = 1

14 = -log[OH-]
=> [OH-] = 10^-14
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