Question

In: Chemistry

How many grams of glyoxylic acid and sodium glyoxylate are needed to prepare 2.50 L of...

How many grams of glyoxylic acid and sodium glyoxylate are needed to prepare 2.50 L of a 1.50 M buffer at pH 4.10? The pKa of glyoxylic acid is 3.34. Note: Use the monohydrate forms, HCOCO2H·H2O and HCOCO2Na·H2O.

Solutions

Expert Solution

Given,

Concentration of buffer = 1.50 M

The volume of buffer = 2.50 L

pH = 4.10

pKa of glyoxylic acid = 3.34

Assume,

x = moles of glyoxylic acid

y = moles of sodium glyoxylate

From the given volume and concentration of buffer, calculating the total number of moles of acid and salt in the buffer,

= 2.50 L x 1.50 M

= 3.75 mol of acid and salt(C.Base)

Thus, x + y = 3.75 ----------(1)

Now, We know, the Henderson-Hasselbalch equation,

pH = pKa + log [ C.Base/ Acid]

4.10 = 3.34 + log [ y/ x]

Thus, y/x = 5.7544

Thus, y = 5.7544 x ---------- (2)

Substituting the value of "y" in equation (1)

x + 5.7544 x = 3.75 ----------(1)

x = 0.5552

Also,

y = 5.7544 x

y = 5.7544 x 0.5552

y = 3.1948

Thus,

Moles of glyoxylic acid = x = 0.5552 mol

Moles of sodium glyoxylate = y = 3.1948 mol

Converting these number of moles to grams,

= 0.5552 mol  glyoxylic acid monohydrate x ( 92.05 g / 1 mol)

= 51.1 g glyoxylic acid monohydrate

= 3.1948 mol sodium glyoxylate monohydrate x (114.032 g/ 1 mol)

= 364.3 g sodium glyoxylate monohydrate

Thus, 51.1 grams of glyoxylic acid and 364 g of sodium glyoxylate are needed to prepare 2.50 L of 1.50 M buffer at pH 4.10.


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