Experiment 12 Pre-Laboratory Assignment 4. If you need to prepare 250.0 mL of a

Question

Experiment 12 Pre-Laboratory Assignment 4. If you need to prepare 250.0 mL of a pH-3.60 buffer that has a total buffer concentra- tion of formic acid+ formate of 0.030 M, how many moles of each will you need? 5. Discuss how to prepare the solution in Question 4 starting from 0.100 M formic acid and 0.150 M sodium formate. 6. You realize that you have spilled a few drops of sulfuric acid solution on your sleeve, but there is no visible harm. Can you be confident that the fabric is safe? What can you do to protect yourself from such mishaps?

Explanation / Answer

Given:

[Formate, HCOO-] + [Formic Acid, HCOOH] = 0.030 M

Volume of the solution = 250 mL = 0.250L

pH of the solution = 3.60

To Find: Moles of Formate and Formic acid required to make the above solution.

Solution:

In order to find the moles of each component, we need molarity and when molarity is multiplied with the given volume, we can get moles.

We know that [Formate, HCOO-] + [Formic acid, HCOOH] = 0.030 M

Let [Formate, HCOO-] = x,

then [Formic acid, HCOOH] = 0.030 - x,

We can now use the Hasslebach henderson equation:

pH = pKa + log [Base/Acid]

pKa of formic acid = 3.74

3.60 = 3.74 + log[HCOO-/HCOOH]

3.60 = 3.74 + log[x/0.03-x]

log[x/0.03-x] = -0.14

[x/0.03-x] = 10-0.14

x/(0.03-x) = 0.724

x = 0.724(0.03-x)

x = 0.724(0.03-x)

x = 0.0217 - 0.724x

1.72 x = 0.0217

x= 0.217/1.72 = 0.0126 M

[Formate, HCOO-] = x =  0.0126 M

moles of Formate = 0.0126 moles/L * 0.250 L = 3.16 x 10-3 ?moles

[Formic acid, HCOOH] = 0.030 - x = 0.030 - 0.0126 = 0.0174

moles of Formic acid = 0.0174 moles/L * 0.250 L = 4.34 x 10-3 moles

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