Respuesta :
Answer:
For 0.1 M sodium acetate solution
if concentration of acid is 0.0025 then pH will 6.075
if concentration of acid is 0.005 then pH will 5.775
if concentration of acid is 0.01 then pH will 5.475
if concentration of acid is 0.05 then pH will 4.775
For 0.01 M sodium acetate solution
if concentration of acid is 0.0025 then pH will 5.075
if concentration of acid is 0.005 then pH will 4.775
if concentration of acid is 0.01 then pH will 4.475
if concentration of acid is 0.05 then pH will 3.775
Explanation:
to calculate the pH of a buffer solution we use the following formula
pH = pKa + log [B]/[A] ------------- eq (1)
[B] = concentration of base
[A] = concentration of acid
Given data
[B] = 0.1 M , 0.01M
[A] = 0.0025 M , 0.005 M, 0.01 M, 0.05 M
pKa value for sodium acetate is 4.75
1. First we will calculate the pH values for 0.1 M acetate solution.
If the concentration of acid is 0.0025, then:
[B] = 0.1 M
[A] = 0.0025 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.0025]
pH = 4.75 + log [40]
pH = 4.475 + 1.6
pH = 6.075
If the concentration of acid is 0.005 M, then:
[B] = 0.1 M
[A] = 0.005 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.005]
pH = 4.75 + log [20]
pH = 4.475 + 1.3
pH = 5.775
If the concentration of acid is 0.01, then:
[B] = 0.1 M
[A] = 0.01 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.01]
pH = 4.75 + log [10]
pH = 4.475 + 1
pH = 5.475
If the concentration of acid is 0.05, then:
[B] = 0.1 M
[A] = 0.05 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.05]
pH = 4.75 + log [2]
pH = 4.475 + 0.3
pH = 4.775
2. Now we will calculate the pH values for 0.01 M acetate solution.
If the concentration of acid is 0.0025, then:
[B] = 0.01 M
[A] = 0.0025 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.0025]
pH = 4.75 + log [4]
pH = 4.475 + 0.6
pH = 5.075
If the concentration of acid is 0.005 M, then:
[B] = 0.01 M
[A] = 0.005 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.005]
pH = 4.75 + log [2]
pH = 4.475 + 0.3
pH = 4.775
If the concentration of acid is 0.01 M, then:
[B] = 0.01 M
[A] = 0.01 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.01]
pH = 4.75 + log [1]
pH = 4.475 + 0
pH = 4.475
If the concentration of acid is 0.05 M, then:
[B] = 0.01 M
[A] = 0.05 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.05]
pH = 4.75 + log [0.2]
pH = 4.475 + (-0.7)
pH = 4.475 - 0.7
pH = 3.775
Answer:
a) pH = 4.71
b) pH = 4.704
c) pH = 4.57
d) No buffer here, the pH will be between 2-3
Explanation:
Applying Henderson Hasselbach equation:
pH = pKa + log([A]/[HA])
a) For 0.0025 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.0025 M of acid:
[A] = 0.05 - 0.0025 = 0.0475 M
[HA] = 0.05 + 0.0025 = 0.0525 M
[tex]pH=4.75+log(\frac{0.0475}{0.0525} )=4.71[/tex]
b) For 0.005 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.005 M of acid:
[A] = 0.05 - 0.005 = 0.0495 M
[HA] = 0.05 + 0.005 = 0.055 M
[tex]pH=4.75+log(\frac{0.0495}{0.055} )=4.704[/tex]
c) For 0.01 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.01 M of acid:
[A] = 0.05 - 0.01 = 0.04 M
[HA] = 0.05 + 0.01 = 0.06 M
[tex]pH=4.75+log(\frac{0.04}{0.06} )=4.57[/tex]
d) For 0.05 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.05 M of acid:
[A] = 0.05 - 0.05 = 0
[HA] = 0.05 + 0.05 = 0.1 M
No buffer here, the pH will be between 2-3
