Question

Using only sodium carbonate, Na2CO3, sodium bicarbonate, NaHCO3, and distilled water determine how you could prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3. The total molarity of the ions should be 0.20 M. The Ka of the hydrogen carbonate ion, HCO3 - , is 4.7 x 10-11 .

Answer 1

Answer:

Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.

Explanation:

The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:

$pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]})$  

We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:

$10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]})$  

$\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94$  (1)

Also, we know that:

$[Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M$    (2)

From equation (2) we have:

$[Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}]$   (3)

By entering (3) into (1):

$\frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94$

$0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20$

$[NaHCO_{3}] = 0.103 M$  

Hence, the [Na_{2}CO_{3}] is:

$[Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M$  

Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:

$m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g$

$m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g$

Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.      

   

I hope it helps you!