Question

In: Chemistry

Malonic acid (H2M) is a diprotic acid. You wish to prepare a buffer of malonic acid...

Malonic acid (H2M) is a diprotic acid. You wish to prepare a buffer of malonic acid (pKa1 = 2.847, pKa2= 5.696) with a final pH of 2.90 but you only have disodium malonate (M2-) in your laboratory shelf. You dissolve 100 mmol of this salt in 1 L of water. Assume both dissociation processes are decoupled.

What is the initial pH of the solution? State any assumptions you make.

How many equivalents of strong acid must you add to your solution to end with only HM- ?

What is the approximate pH of this solution of HM- ? Use an adequate mathematical approximation.

How many additional equivalents must you add to reach the desired pH of 2.90?

Expert Solution

What is the initial pH of the solution? State any assumptions you make.

Initial [M^2-] = 0.1 mol/L (C_base)

Initial pH:

The simplest calculation

pOH = 0.5pKb -0.5logCbase

= 0.5(14-5.696) -0.5log0.1

= 4.152+0.5 = 4.652

Note. 10^-4.652 << C_base (the simple calculation is justified)

Initial pH = 14 – 4.652 = 9.348

How many equivalents of strong acid must you add to your solution to end with only HM- ?

100 mequiv

What is the approximate pH of this solution of HM- ? Use an adequate mathematical approximation.

HM- is an amphoter

pH = (pKa1+pka2)/2 = 4.271

How many additional equivalents must you add to reach the desired pH of 2.90?

Buffer pH

pka1 < buffer pH < pKa2 and buffer pH-pKa1<1. Then

pH = pKa1 + log ([HM^-]/[H2M])

2.90 = 2.847 + log([HM^-]/[H2M])

[HM^-]/[H2M] = 0.053:1

[HM^-] / C_{total H2M} = 0.053/1.053 = 0.050

In the second neutralization step, the addition of acid have to be only

100 x (1 – 0.050) = 95% of the quantity needed for a complete neutralization.

Answer: 95% x 100 mequiv.= 95 mequiv.

queen_honey_blossom answered 1 year ago

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