a) You have to prepare a pH 5.25 buffer from the following 0.100 M solutions available:...

a) You have to prepare a pH 5.25 buffer from the following 0.100 M solutions available:

1. hydrocyanic acid

2. hypobromous acid

3. propanoic acid

4. potassium cyanide

5. potassium hypobromite

6. potassium propanoate

Which solutions would you choose to make the most effective buffer? Why?

b) How many milliliters of each of each o the solutions selected in part a would you use to make one liter of the final buffer solution?

Solutions

Expert Solution

a.)

Propanoic acid - Ka = 1.3 x 10^–⁵

Of the acids listed, the K_a value for propanoic acid to potassium propanoate is closest to the desired hydrogen ion concentration. Therefore, you need only to adjust the ratio of [HC₃H₅O₂]/[HC₃H4O₂^-] to get the desired final hydrogen ion concentration. You can do this using the Henderson-Hasselbalch equation:

So, 3 and 6 should be used.

Required buffer,

pH = 5.25

H+ = 5.6 X 10^-6

pH = pKa + log [acid]/[base]

H+ = Ka([HC₃H₅O₂]/[HC₃H4O₂^-])

5.6 X 10^{-6 =} 1.3 x 10^–5([HC₃H₅O₂]/[HC₃H4O₂^-])

([HC₃H₅O₂]/[HC₃H4O₂^-]) = 5.6/13 = 0.43/1

To satisfy the expression, the ratio of [HC₃H₅O₂]/[HC₃H4O₂^-] must be 0.43 to 1. Therefore, if you add 0.43 mol propanoic acid acid and 1.0 mol ptassium propanoate (or any other pair of amounts such that the ratio is still 2.8 to 1) to enough water to make 1.0 L of solution, the solution will be a buffer with a pH of 5.25.

So,, we have to use 4.3 L of 3.propanoic acid and 10 L of potassium propanoate to obtain buffer of pH =5.25.

queen_honey_blossom answered 1 year ago

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