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guapka [62]
3 years ago
11

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?

Chemistry
1 answer:
tester [92]3 years ago
7 0

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!

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