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 ![\frac {[conjugate \ base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5Bconjugate%20%5C%20base%5D%7D%7B%5Bacid%5D%7D)
6.65 = 7.21 + log ![\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO_4%5E%7B2-%7D%5D%7D%7B%5BH_2PO_%7B4%7D%5E-%5D%7D)
6.65 - 7.21 = log ![\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO_4%5E%7B2-%7D%5D%7D%7B%5BH_2PO_%7B4%7D%5E-%5D%7D)
-0.56 = log ![\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO_4%5E%7B2-%7D%5D%7D%7B%5BH_2PO_%7B4%7D%5E-%5D%7D)
= ![10^{-0.56}](https://tex.z-dn.net/?f=10%5E%7B-0.56%7D)
= 0.275
From the question ; given that the conjugate base = 1 M
i.e [HPO₄²⁻] = 1 M
Then;
![\frac{1 \ M }{[H_2PO_4^-]} = 0.275](https://tex.z-dn.net/?f=%5Cfrac%7B1%20%5C%20M%20%7D%7B%5BH_2PO_4%5E-%5D%7D%20%3D%200.275)
![[H_2PO_4^-] =\ \frac{1 \ M } {0.275}](https://tex.z-dn.net/?f=%5BH_2PO_4%5E-%5D%20%3D%5C%20%5Cfrac%7B1%20%5C%20M%20%7D%20%7B0.275%7D)
![[H_2PO_4^-] =\ 3.63 \ M](https://tex.z-dn.net/?f=%5BH_2PO_4%5E-%5D%20%3D%5C%203.63%20%5C%20M)
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 = ![\frac{mass \ ( in \ gram )}{Molar \ mass} * \frac{1}{volume \ in \ litre }](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%20%5C%20%28%20in%20%20%5C%20gram%20%29%7D%7BMolar%20%5C%20mass%7D%20%2A%20%5Cfrac%7B1%7D%7Bvolume%20%5C%20in%20%5C%20litre%20%7D)
1 M = ![\frac{mass \ ( in \ gram )}{141.96} * \frac{1}{1.00 \ L}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%20%5C%20%28%20in%20%20%5C%20gram%20%29%7D%7B141.96%7D%20%2A%20%5Cfrac%7B1%7D%7B1.00%20%5C%20L%7D)
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 = ![\frac{mass \ ( in \ gram )}{Molar \ mass} * \frac{1}{volume \ in \ litre }](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%20%5C%20%28%20in%20%20%5C%20gram%20%29%7D%7BMolar%20%5C%20mass%7D%20%2A%20%5Cfrac%7B1%7D%7Bvolume%20%5C%20in%20%5C%20litre%20%7D)
3.63 M = ![\frac{mass \ ( in \ gram )}{119.98} * \frac{1}{1.00 \ L}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%20%5C%20%28%20in%20%20%5C%20gram%20%29%7D%7B119.98%7D%20%2A%20%5Cfrac%7B1%7D%7B1.00%20%5C%20L%7D)
mass (in gram ) = 3.63 × 119.98
= 435.52 g
Mass ( in gram ) for sodium salt of the weak acid = 435.52 g