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:
We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:
(1)
Also, we know that:
(2)
From equation (2) we have:
(3)
By entering (3) into (1):
![0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20](https://tex.z-dn.net/?f=%200.94%2A%5BNaHCO_%7B3%7D%5D%20%2B%20%5BNaHCO_%7B3%7D%5D%20%3D%200.20%20)
Hence, the [Na_{2}CO_{3}] is:
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](https://tex.z-dn.net/?f=m_%7BNa_%7B2%7DCO_%7B3%7D%7D%20%3D%20C%2AV%2AM%20%3D%200.097%20mol%2FL%2A0.050%20L%2A105.99%20g%2Fmol%20%3D%200.5141%20g)
![m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g](https://tex.z-dn.net/?f=m_%7BNaHCO_%7B3%7D%7D%20%3D%20C%2AV%2AM%20%3D%200.103%20mol%2FL%2A0.050%20L%2A84.007%20g%2Fmol%20%3D%200.4326%20g)
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!