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.

We need 4.28 grams of sodium formate

Explanation:

Step 1: Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

Step 2: Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

Step 3: Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

Step 4: Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate