Answer:
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] <em>(1)</em>
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
<em>2,30 moles = [A⁻] </em>
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) =<em> 156,4 g of sodium formate</em>.
I hope it helps!
