Answer :
The pH of the solution is approximately 13.47.
To find the pH of the solution, we first need to determine the concentrations of the carbonate (CO3^2-) and bicarbonate [tex](HCO3^-)[/tex] ions in the solution, which result from the dissociation of Na2CO3 and NaHCO3, respectively.
Calculate the moles of Na2CO3 and NaHCO3:
Moles of Na2CO3[tex]: \( \frac{5.00 \text{ g}}{105.99 \text{ g/mol}} \) (molar mass of Na2CO3 is 105.99 g/mol)[/tex]
Moles of NaHCO3:[tex]\( \frac{5.00 \text{ g}}{84.01 \text{ g/mol}} \) (molar mass of NaHCO3 is 84.01 g/mol)[/tex]
[tex](CO3^2-)[/tex]
Calculate the total volume of the solution: 0.100 L
Calculate the concentration of each ion:
- For Na2CO3: Since it dissociates into 2 moles of [tex]Na^+ ions[/tex] and 1 mole of [tex]CO3^2[/tex]- ions, the concentration of [tex]CO3^2[/tex]- ions is[tex]\( \frac{2 \times \text{moles of Na2CO3}}{\text{total volume of solution}} \).[/tex]
- For NaHCO3: Since it dissociates into 1 mole of Na^+ ions and 1 mole of [tex]HCO3^-[/tex] ions, the concentration of [tex]HCO3^-[/tex] ions is[tex]\( \frac{\text{moles of NaHCO3}}{\text{total volume of solution}} \).[/tex]
Calculate the [tex]\( \text{pOH} \)[/tex] of the solution:
[tex]\( \text{pOH} = -\log{[\text{OH}^-]} \)[/tex]
Since [tex]\( \text{pOH} = -\log{[\text{CO3}^{2-}]} \) for Na2CO3 and \( \text{pOH} = -\log{[\text{HCO3}^-]} \) for NaHCO3[/tex], we'll calculate the pOH for each and then add them together.
Calculate the pH of the solution:
[tex]\( \text{pH} = 14 - \text{pOH} \)[/tex]
Let's go through these steps:
Calculate moles:
Moles of Na2CO3: [tex]\( \frac{5.00 \text{ g}}{105.99 \text{ g/mol}} \) ≈ 0.0471[/tex] mol
Moles of NaHCO3: [tex]\( \frac{5.00 \text{ g}}{84.01 \text{ g/mol}} \) ≈ 0.0595[/tex]mol
Calculate total volume:
[tex]\( \text{Total volume} = 0.100 \text{ L} \)[/tex]
Calculate concentrations:
[tex]\( [\text{CO3}^{2-}] = \frac{2 \times 0.0471 \text{ mol}}{0.100 \text{ L}} \)\\ \( [\text{HCO3}^-] = \frac{0.0595 \text{ mol}}{0.100 \text{ L}} \)[/tex]
Calculate [tex]\( \text{pOH} \):[/tex]
[tex]\( \text{pOH} = -\log{[\text{CO3}^{2-}]} - \log{[\text{HCO3}^-]} \)[/tex]
Calculate [tex]\( \text{pH} \):[/tex]
[tex]\( \text{pH} = 14 - \text{pOH} \)[/tex]
Let's do the calculations:
. Moles of Na2CO3
[tex]\( \frac{5.00 \text{ g}}{105.99 \text{ g/mol}} \) ≈ 0.0471 mol[/tex]
Moles of NaHCO3:
[tex]\( \frac{5.00 \text{ g}}{84.01 \text{ g/mol}} \) ≈ 0.0595 mol[/tex]
Total volume:
0.100 L
Concentrations:
[tex]\( [\text{CO3}^{2-}] = \frac{2 \times 0.0471 \text{ mol}}{0.100 \text{ L}} \) ≈ 0.942 M\\ \( [\text{HCO3}^-] = \frac{0.0595 \text{ mol}}{0.100 \text{ L}} \) ≈ 0.595 M\\[/tex]
[tex]\( \text{pOH} = -\log{0.942} - \log{0.595} \) ≈ 0.526[/tex]
[tex]\( \text{pH} = 14 - 0.526 \) ≈ 13.47[/tex]
So, the pH of the solution is approximately 13.47.
