What weight of sodium formate must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50?

156,4 g of sodium formate

Explanation:

The pka of the formic acid is 3,74. Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] (1)

Where A⁻ is the conjugate base (Formate) and HA is the formic acid.

4.00L of 1.00M formic acid contain:

4.00L × (1.00mol /L) = 4.00 moles

Replacing these moles, the desired pH and the pka value in (1):

3,50 = 3,74  log₁₀ [A⁻] / 4,00 moles

-0,24 =  log₁₀ [A⁻] / 4,00 moles

0,575 = [A⁻] / 4,00 moles

2,30 moles = [A⁻]

That means you need 2,30 moles of formate (Sodium formate), to produce the desired buffer. As the molar mass of sodium formate is 68,01g/mol, the weight of sodium formate that must be added is:

2,30mol×(68,01g/mol) = 156,4 g of sodium formate.

I hope it helps!

aluminum

i guess but i think im right