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

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.

Step-by-step explanation:

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

`pH = pKa + log(([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(([Na_(2)CO_(3)])/([NaHCO_(3)]))`

`([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):

`(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!