Design a buffer that has a pH of 6.65 using one of the weak acid/ conjugate base systems shown below
Weak Acid Conjugate Base Ka pKa
HC₂O₄⁻ C₂O₄² ⁻ 6.4 × 10⁻⁵ 4.19
H₂PO₄⁻ HPO₄²⁻ 6.2 × 10⁻⁸ 7.21
HCO₃⁻ CO₃² ⁻ 4.8 × 10⁻¹¹ 10.32
How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?
a) grams sodium salt of the conjugate base = <u>????? g </u>
b) grams sodium salt of weak acid = <u>????? g </u>
Answer:
a) 141.96 g
b) 435.52 g
Explanation:
From the question ; in order to design a buffer solution that has a pH of 6.65; we need to take a close look at the system that will eventually have a pKa close to 6.65
From the given question; we see that H₂PO₄⁻/ HPO₄²⁻ is best fit for the process .
Now, using Henderson - Hasselbalch equation
pH = pKa + log
6.65 = 7.21 + log
6.65 - 7.21 = log
-0.56 = log
=
= 0.275
From the question ; given that the conjugate base = 1 M
i.e [HPO₄²⁻] = 1 M
Then;
To determine the mass ( in gram ) of the required salts ; we have:
[HPO₄²⁻] = 1 M
[ Na₂HPO₄] = 1 M
Molar mass of Na₂HPO₄ in the conjugate base = 141.96 g/mol
Molarity =
1 M =
mass (in gram ) = 1 × 141.96
= 141.96 g
Mass ( in gram ) for sodium salt of the conjugate base = 141.96 g
b)
Also, gram sodium salt of weak acid can be calculated as follows:
[H₂PO₄⁻] = 3.63 M
[ Na₂HPO₄] = 3.63 M
Molar mass of Na₂HPO₄ in the weak acid = 119.98 g/mol
Molarity =
3.63 M =
mass (in gram ) = 3.63 × 119.98
= 435.52 g
Mass ( in gram ) for sodium salt of the weak acid = 435.52 g