Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
Explanation:
Given: Total volume of the buffer = 21.0 mL
[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)
It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.
Hence,
[HCOONa] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] x
= 0.1x mmol
Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] (21.0 - x) mmol
Using equation (1),
[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]
As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.
Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
