Explanation:
Given:
Ka = 1.8 × 10-4
HCOOH --> HCOO- + H+
HA --> H+ + A-
Ka = [H+] × [A-]/[HA]
[HA] = 0.12 M
[H+] = [A-] = x^2
1.8 × 10^-4 × 0.12 = x^2
x = 0.00465 M
pH = - log [H+]
= -log [0.00465]
= 2.33
B.
Since 1 mole of NaOH reactes with 1 mole of HCOOH.
Number of moles of NaOH = 0.08 × 15 × 10^-3
= 0.0012 mole.
Concentration of HCOOH = 0.0012/20 × 10^-3
= 0.06 M
[HA] = 0.06 M
1.8 × 10^-4 × 0.06 = x^2
x = 0.00329 M
pH = - log [H+]
= -log [0.00329]
= 2.48
C.
Since 1 mole of NaOH reacts with 1 mole of HCOOH.
Number of moles of NaOH = 0.08 × 30 × 10^-3
= 0.0024 mole.
Concentration of HCOOH = 0.0024/20 × 10^-3
= 0.12 M
[HA] = 0.12 M
1.8 × 10^-4 × 0.12 = x^2
x = 0.00465 M
pH = - log [H+]
= -log [0.00465]
= 2.33
D.
Since 1 mole of NaOH reactes with 1 mole of HCOOH.
Number of moles of NaOH = 0.08 × 30.5 × 10^-3
= 0.00244 mole.
Concentration of HCOOH = 0.00244/20 × 10^-3
= 0.122 M
[HA] = 0.122 M
1.8 × 10^-4 × 0.122 = x^2
x = 0.00469 M
pH = - log [H+]
= -log [0.00469]
= 2.33
