Acid/Base Equilibria pH?
If someone could tell me how to do these sorts of problems, it would be great!
A 0.200 M solution of a weak base in water has a pH = 10.40 at 25 C. Calculate the value of Kb for this base.
I know that pH = pKa + log[A/HA] but I cant figure out the answer.
Formic acid, HCO2H, has an ionization constant with the value: Ka= 1.76 x 10^-4. Calculate the value of pKb for the conjugate base of formic acid.
I don't know how to get from Ka to Kb. I know that Ka x Kb = Kw
The ionization constant, Ka, for benzoic acid, HC7H502, is 6.28 x 10^-5 . What is he pH of a 0.15 molar solution of this acid?
Thanks to whoever can help me! Greatly appreciated.
Comments
A 0.200 M solution of a weak base in water has a pH = 10.40 at 25 C. Calculate the value of Kb for this base.
I know that pH = pKa + log[A/HA] but I cant figure out the answer.
Answer: You do not have a buffer solution therefore you cannot use the Henderson - Hasselbalch equation that you quote:
If pH = 10.40 then
pOH = 14.00-10.40 = 3.60
[OH-] = 10^-pOH
[OH-] = 10^-3.60
Use Kb equation
Kb = [OH-]² / [base]
Kb = (10^-3.6)² / 0.200
Kb = 6.31*10^-8 / 0.200
Kb = 3.15*10^-7
Formic acid, HCO2H, has an ionization constant with the value: Ka= 1.76 x 10^-4. Calculate the value of pKb for the conjugate base of formic acid.
I don't know how to get from Ka to Kb. I know that Ka x Kb = Kw
You know that Ka * Kb = 10^-14
Kb = 10^-14 / Ka
Kb = 10^-14 / ( 1.76*10^-4)
Kb = 5.68*10^-11
pKb = - log Kb
pKb = -log ( 5.68*10^-11)
pKb = 10.25
The ionization constant, Ka, for benzoic acid, HC7H502, is 6.28 x 10^-5 . What is he pH of a 0.15 molar solution of this acid?
Use Ka equation to calculate [H+]
Equation:
Ka = [H+]² / [acid]
6.28*10^-5 = [H+]² / 0.15
[H+]² = (6.28*10^-5)* 0.15
[H+]² = 9.42*10^-6
[H+] = 3.07*10^-3
pH = -log [H+]
pH = -log ( 3.07*10^-3)
pH = 2.51