In: Chemistry
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.
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.