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] (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!
The weight of sodium formate that must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50 is 156.4g.
The number of moles can be calculated using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where;
4.00L × 1.0M = 4.00 moles
Replacing these moles, the desired pH and the pka value in the above equation;
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⁻]
If 2.30 moles of formate, the mass of formate can be calculated as follows:
2,30mol × 68.01g/mol = 156.4 g of sodium formate
Therefore, the weight of sodium formate that must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50 is 156.4g.
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