Answer :
To solve this problem, we want to find out how many milliliters (mL) of 2.00 M NaOH are needed to titrate a 50.0 mL sample of a 1.00 M solution of the diprotic acid [tex]\(H_2A\)[/tex] to reach a pH of 12.
To do this, we need to follow these steps:
1. Determine the Moles of [tex]\(H_2A\)[/tex]:
- Calculate the initial moles of [tex]\(H_2A\)[/tex] in the solution:
[tex]\[
\text{Moles of } H_2A = \text{Volume (L)} \times \text{Molarity (M)} = \frac{50.0}{1000} \times 1.00 = 0.05 \text{ moles}
\][/tex]
2. Neutralization Process:
- Since [tex]\(H_2A\)[/tex] is a diprotic acid, it can donate two protons. Therefore, it will react with NaOH in a 1:2 ratio (1 mole of [tex]\(H_2A\)[/tex] reacts with 2 moles of NaOH).
- Therefore, the moles of NaOH required for complete neutralization will be:
[tex]\[
\text{Moles of NaOH for complete neutralization} = 2 \times 0.05 = 0.1 \text{ moles}
\][/tex]
3. Calculate the [tex]\(OH^-\)[/tex] Concentration for pH 12:
- The relationship between pH and [tex]\(OH^-\)[/tex] concentration at pH 12 is found by knowing pOH:
[tex]\[
\text{pOH} = 14 - \text{pH} = 14 - 12 = 2
\][/tex]
- Therefore, the concentration of [tex]\(OH^-\)[/tex] needed is:
[tex]\[
[OH^-] = 10^{-2} = 0.01 \text{ M}
\][/tex]
4. Determine Volume Changes and Additional [tex]\(OH^-\)[/tex] Needed:
- The volume of the final solution is roughly estimated to be the sum of the initial solution and the volume of NaOH added. For this calculation, it's simplified to assume:
[tex]\[
V_{\text{final}} \approx V_{H_2A} + V_{H_2A} = 2 \times 50.0 = 100.0 \text{ mL}
\][/tex]
- Calculate the additional moles of [tex]\(OH^-\)[/tex] required to achieve the desired [tex]\(OH^-\)[/tex] concentration in the final solution:
[tex]\[
\text{Excess moles of } OH^- = 0.01 \times \frac{100}{1000} = 0.001 \text{ moles}
\][/tex]
5. Total NaOH Moles and Volume Required:
- The total moles of NaOH required will be the sum of the moles needed for neutralization and the excess moles needed for achieving pH 12:
[tex]\[
\text{Total moles of NaOH} = 0.1 + 0.001 = 0.101 \text{ moles}
\][/tex]
- Finally, calculate the volume of NaOH needed using its molarity:
[tex]\[
V_{\text{NaOH}} = \frac{0.101}{2.00} \times 1000 = 50.5 \text{ mL}
\][/tex]
Therefore, you need to add 50.5 mL of 2.00 M NaOH to reach a pH of 12 in this solution.
To do this, we need to follow these steps:
1. Determine the Moles of [tex]\(H_2A\)[/tex]:
- Calculate the initial moles of [tex]\(H_2A\)[/tex] in the solution:
[tex]\[
\text{Moles of } H_2A = \text{Volume (L)} \times \text{Molarity (M)} = \frac{50.0}{1000} \times 1.00 = 0.05 \text{ moles}
\][/tex]
2. Neutralization Process:
- Since [tex]\(H_2A\)[/tex] is a diprotic acid, it can donate two protons. Therefore, it will react with NaOH in a 1:2 ratio (1 mole of [tex]\(H_2A\)[/tex] reacts with 2 moles of NaOH).
- Therefore, the moles of NaOH required for complete neutralization will be:
[tex]\[
\text{Moles of NaOH for complete neutralization} = 2 \times 0.05 = 0.1 \text{ moles}
\][/tex]
3. Calculate the [tex]\(OH^-\)[/tex] Concentration for pH 12:
- The relationship between pH and [tex]\(OH^-\)[/tex] concentration at pH 12 is found by knowing pOH:
[tex]\[
\text{pOH} = 14 - \text{pH} = 14 - 12 = 2
\][/tex]
- Therefore, the concentration of [tex]\(OH^-\)[/tex] needed is:
[tex]\[
[OH^-] = 10^{-2} = 0.01 \text{ M}
\][/tex]
4. Determine Volume Changes and Additional [tex]\(OH^-\)[/tex] Needed:
- The volume of the final solution is roughly estimated to be the sum of the initial solution and the volume of NaOH added. For this calculation, it's simplified to assume:
[tex]\[
V_{\text{final}} \approx V_{H_2A} + V_{H_2A} = 2 \times 50.0 = 100.0 \text{ mL}
\][/tex]
- Calculate the additional moles of [tex]\(OH^-\)[/tex] required to achieve the desired [tex]\(OH^-\)[/tex] concentration in the final solution:
[tex]\[
\text{Excess moles of } OH^- = 0.01 \times \frac{100}{1000} = 0.001 \text{ moles}
\][/tex]
5. Total NaOH Moles and Volume Required:
- The total moles of NaOH required will be the sum of the moles needed for neutralization and the excess moles needed for achieving pH 12:
[tex]\[
\text{Total moles of NaOH} = 0.1 + 0.001 = 0.101 \text{ moles}
\][/tex]
- Finally, calculate the volume of NaOH needed using its molarity:
[tex]\[
V_{\text{NaOH}} = \frac{0.101}{2.00} \times 1000 = 50.5 \text{ mL}
\][/tex]
Therefore, you need to add 50.5 mL of 2.00 M NaOH to reach a pH of 12 in this solution.
