Calculate the pH of each of the solutions and the change in pH to 0.01 pH units
ID: 907154 • 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 2.05-M HCl to 570. mL of each of the following solutions.
a) water
before mixing = 7
After mixing = 1.45
b) 0.135 M C2H3O21-
Before mixing = 8.95
After mixing = ?
c) 0.135 M HC2H3O2
Before mixing = ?
After mixing = ?
d) a buffer solution that is 0.135 M in each C2H3O21- and HC2H3O2
Before mixing = ?
After mixing = ?
I've tried this so many times and I don't understand it. Is there a formula or a set of formulas for this ?? Can someone please just spell it out for me step by step and why. Please and thank you!
Explanation / Answer
a) water
Before mixing: pH = 7.00
After mixing: [H+] is provided by HCl:
[H+] = 10*2.05/580 = 0.035
Hence: pH = log[H+] = 1.45
pH change = 7-1.45 = 5.55
b) 0.135 M C2H3O21- :
Before mixing:
C2H3O21- + H2O <==> CH3COOH + OH-
Kb = [CH3COOH]*[ OH-]/[ C2H3O21-] = [CH3COOH]*[ OH-]*[H+]/[ C2H3O21-][H+] = Kw/Ka
Notice here CH3COOH and OH- are generated in pairs, hence [CH3COOH] = [OH-], and
Kb = Kw/Ka = [CH3COOH]*[ OH-]/[ C2H3O21-] = [OH-]^2/C
[OH-] = (C*Kw/Ka)
[H+] = Kw/[OH-] = ( Kw.Ka/C)
log([H+]) = -log( Kw.Ka/C)
= 0.5(logKw + logKa-logC)
pH = 0.5*(pKw +pKa +logC)
=0.5(14+4.76+log0.135)
= 8.95
After mixing:
The [H+] concentration provided by HCl in mixing is:
10*2.05/580=0.035M
The [C2H3O21- ] concentration in mixing is:
0.135M*(570/580) = 0.133M
The reaction between H+ and C2H3O21- forms CH3COOH and Cl-,
[CH3COOH] = 0.035M with [C2H3O21-] = 0.133-0.035=0.098M left.
Kb = 5.75 X 10 -10= [CH3COOH]*[ OH-]/[ C2H3O21-] = [OH-]*0.035/0.098
[OH-] = 1.61*10-9
pOH = 8.79
pH = 14-8.79 =5.21
c) 0.135 M HC2H3O2:
Before mixing:
CH3COOH + H2O = CH3COO- + H3O+
Ka = [CH3COO-] [H3O+]/ [CH3COOH]
=C.C /(1-)C....[ is degree of dissociation]
= C2 ...(As it is a weak acid, dissociation is very little, is very small)
= (Ka/C)
[H3O+] = C = C* (Ka/C) = (Ka.C)
= (1.73*10-5*0.135
=1.53*10-3
pH=2.81
After mixing:
Notice CH3COOH is a weak acid and therefore (1) there is no reaction between CH3COOH and HCl, and HCl provide more [H+] and CH3COOH dissociate much less.
The HCl or [H+] concentration after mixing is:
10*2.05/580=0.035M
pH = 1.45
d. a buffer solution that is 0.135 M in each C2H3O21- and HC2H3O2:
Before mixing: The [C2H3O21- ] concentration or [CH3COOH] in mixing is:
0.135M*(570/580) = 0.133M
Apply Henderson equation for buffer solution,
pH = pKa + log [salt]/[acid]
= 4.76+ log0.133/0.133
=4.76
After mixing:
C2H3O21- + H+ <==> CH3COOH
This concentration of salt decreases and concentration of acid increases.
[H+] is provided by HCl:
[H+] = 10*2.05/580 = 0.035
The [C2H3O21- ] concentration or [CH3COOH] in mixing is:
0.135M*(570/580) = 0.133M
After reaction, [C2H3O21- ] = 0.133-0.035=0.098 M
[CH3COOH] = 0.133+0.015 = 0.148 M
Apply Henderson equation for buffer solution,
pH = pKa + log [salt]/[acid]
= 4.76+ log0.098/0.148
=4.58
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.