Question

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

Answer 1

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 = ?????  g    

b)  grams sodium salt of weak acid =  ?????  g    

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 $\frac {[conjugate \ base]}{[acid]}$

6.65 = 7.21   + log $\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}$

6.65 - 7.21 = log $\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}$

-0.56 = log $\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}$

$\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}$ = $10^{-0.56}$

$\frac{[HPO_4^{2-}]}{[H_2PO_{4}^-]}$ = 0.275

From the question ; given that the conjugate base = 1 M

i.e  [HPO₄²⁻] = 1 M

Then;

$\frac{1 \ M }{[H_2PO_4^-]} = 0.275$

$[H_2PO_4^-] =\ \frac{1 \ M } {0.275}$

$[H_2PO_4^-] =\ 3.63 \ M$

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 = $\frac{mass \ ( in \ gram )}{Molar \ mass} * \frac{1}{volume \ in \ litre }$

1 M = $\frac{mass \ ( in \ gram )}{141.96} * \frac{1}{1.00 \ L}$

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 = $\frac{mass \ ( in \ gram )}{Molar \ mass} * \frac{1}{volume \ in \ litre }$

3.63 M = $\frac{mass \ ( in \ gram )}{119.98} * \frac{1}{1.00 \ L}$

mass (in gram ) = 3.63  × 119.98

= 435.52 g

Mass ( in gram ) for sodium salt of the weak acid = 435.52 g