Calculate the pH of each of the solutions and the change in pH to 0.01 pH units
ID: 855384 • Letter: C
Question
Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.09-M HCl to 510. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.
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Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.09-M HCl to 510. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative a) water pH before mixing = 7.00 pH after mixing= -1.387 pH change = -5.613 b) 0.135 M C2H3O2^1- pH before mixing = 8.94 pH after mixing= 3.575 pH change = 3.88 c) 0.135 M HC2H3O2 pH before mixing = pH after mixing= pH change = d) a buffer solution that is 0.135 M in each C2H3O2^1- and HC2H3O2 pH before mixing = pH after mixing= pH change =Explanation / Answer
Hi,
Just now i have seen this question.very few minutes to left answer.So i am sending one of the example. just change the values.
adding 10.0 mL of 2.22-M HCl to 650. mL of each of the following solutions.
0.0100 litres @ 2.22mol/litre = 0.0222 moles of H+ ion added
=============
a) water
0.0222 moles of H+/ 0.650 litres = 0.0342 Molar H+
pH = 1.47
pH before mixing = 7.00
pH after mixing= 1.47
pH change = -5.33
================
b) 0.186 M C2H3O21-
before Kb of acetate = Kw / Ka acetic acid = 1e-14 / 1.8e-5
Kb = 5.56 e-10 = [acetic acid ] [OH-] /[acetate]
5.56 e-10 = [X ] [X] /[0.186]
X = [OH-] = 1.02e-5
pOH = 4.99
pH = 9.01
--------------------------------
adding 0.0342 Molar H+ turns 0.0342 Molar of the acetate into acetic acid
after mixing
H C2H3O2 --> H+ & C2H3O2-
[0.0342] ----> [X] & 0.[152]
Ka = 1.8e-5 = [H+] [C2H3O2] / [acid]
1.8e-5 = [H+] [0.152] / [0.0342]
H+ = 4.05e-6
pH = 5.39
pH before mixing = 8.99
pH after mixing= 5.39
pH change = -3.60
====================
c) 0.186 M HC2H3O2
before
Ka = 1.8e-5 = [H+] [C2H3O2] / [acid]
1.8e-5 = [x] [x] / [0.186]
H+ = 1.83e-3
pH = 2.74
pH before mixing = 2.74
pH after mixing= 1.47 (same as (a)
pH change = -1.27
=============
d) a buffer solution that is 0.186 M in each C2H3O21- and HC2H3O2
afterwards, th equilibrium shifts to the left
HC2H3O2 <--- H+ & C2H3O21-
[0.186 + 0.0342] <-- H+ & [ 0.186 - 0.0342]
Ka = 1.8e-5 = [H+] [0.152] / [0.220]
H+ = 2.61e-5
pH = 4.58
pH before mixing = pKa = 4.74
pH after mixing= 4.58
pH change = - 0.16
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