What mass of solid sodium formate (of mw 68.01) must be added to 150 ml of 0.42 mol/l formic acid (hcooh) to make a buffer solu- tion having a ph of 3.74? ka = 0.00018 for hcooh.
answer in units of g.

4·234 g

Explanation:

Given

Volume of the solution = 150 mL

Concentration of formic acid = 0·42 mol/L

pH of the buffer solution = 3·74

for HCOOH = 0·00018

∴ of weak acid = -log() = -log(0·00018) = 3·745

Let the mass of solid sodium formate that must be added be m g

∴ The concentration of sodium formate = ((m ÷ 68·01) ÷ 150) × 10000

In this case conjugate base is sodium formate and weak acid is formic acid

3·74 = 3·745 + log(((m ÷ 68·01) ÷ 150) × 1000) - log(0·42)

⇒ -0·005 +log(0·42) = log(((m ÷ 68·01) ÷ 150) × 1000)

⇒ -0·005 - 0·377 = log(((m ÷ 68·01) ÷ 150) × 1000)

⇒ -0·382 = log(((m ÷ 68·01) ÷ 150) × 1000)

⇒ antilog(-0·382) = ((m ÷ 68·01) ÷ 150) × 1000

⇒ 0·415 × = ((m ÷ 68·01) ÷ 150

⇒ 0·415 × × 150 × 68·01 = m

∴ m = 4·234 g

∴ Mass of solid sodium formate that must be added is 4·234 g

it is a good conductor

explanation:

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