(e) An acidic buffer solution is formed by mixing 50 mmol dm formic acid (HCOOH) with 0.8 mol dm³ of sodlum formate (HCOONa). The acid dissociation constant (K.) of formic acid is 1.78 × 104, (1) Calculate the pk, of formic acid, (1) Calculate the pH of the buffer solution described above, using the Henderson- Hasselbalch equation: pH = pk + log10 [Conjugate Base] (Acid) (III) A small amount of sodium hydroxide (NaOH) is added to the solution, but the pH does not change, Explain this observation.
(e) An acidic buffer solution is formed by mixing 50 mmol dm formic acid (HCOOH) with 0.8 mol dm³ of sodlum formate (HCOONa). The acid dissociation constant (K.) of formic acid is 1.78 × 104, (1) Calculate the pk, of formic acid, (1) Calculate the pH of the buffer solution described above, using the Henderson- Hasselbalch equation: pH = pk + log10 [Conjugate Base] (Acid) (III) A small amount of sodium hydroxide (NaOH) is added to the solution, but the pH does not change, Explain this observation.
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![(e) An acidic buffer solution is formed by mixing 50 mmol dm formic acid (HCOOH)
with 0.8 mol dm3 of sodium formate (HCOONa). The acid dissociation constant (K.)
of formic acid is 1.78 × 104,
(1) Calculate the pk, of formic acid,
(1)
Calculate the pH of the buffer solution described above, using the Henderson-
Hasselbalch equation!
[Conjugate Base]
(Acid]
pH = pK₂ + log₁0
(III) A small amount of sodium hydroxide (NaOH) is added to the solution, but the
pH does not change, Explain this observation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8a9802b-ded9-4a7b-93c8-71218002814a%2F5db0c5c3-df58-46eb-bdf5-f6a4dbd537d9%2Feax00tf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(e) An acidic buffer solution is formed by mixing 50 mmol dm formic acid (HCOOH)
with 0.8 mol dm3 of sodium formate (HCOONa). The acid dissociation constant (K.)
of formic acid is 1.78 × 104,
(1) Calculate the pk, of formic acid,
(1)
Calculate the pH of the buffer solution described above, using the Henderson-
Hasselbalch equation!
[Conjugate Base]
(Acid]
pH = pK₂ + log₁0
(III) A small amount of sodium hydroxide (NaOH) is added to the solution, but the
pH does not change, Explain this observation.
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