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Suppose you need to prepare 21.0 mL of formate buffer with a ratio of 4 of [sodium formate]/[formic acid] by mixing 0.10 M formic acid and 0.10 M sodium formate. How many milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid)

Respuesta :

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).

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