A 50.0 mL sample of a 1.00 M solution of a diprotic acid `H_(2)A(K_(a1)=1.0xx10^(-6) " and " K_(a2)=1.0xx10^(-10))` is titrated with 2.00 M NaOH. What is the minimum volume of 2.00 M NaOH needed to reach a pH of 10.00 ?

Options

A. 12.5 mL
B. 37.5 mL
C. 25.0 mL
D. 50.0 mL

Answer

To solve the problem of determining the minimum volume of 2.00 M NaOH needed to reach a pH of 10.00 when titrating a 50.0 mL sample of a 1.00 M diprotic acid \( H_2A \) (with \( K_{a1} = 1.0 \times 10^{-6} \) and \( K_{a2} = 1.0 \times 10^{-10} \)), we can follow these steps:

### Step 1: Determine the concentration of \( H^+ \) ions at pH 10.00
The pH is given as 10.00. We can calculate the concentration of \( OH^- \) ions using the formula:
\[ pOH = 14 - pH \]
\[ pOH = 14 - 10 = 4 \]
Now, we find the concentration of \( OH^- \):
\[ [OH^-] = 10^{-pOH} = 10^{-4} \, \text{M} \]

### Step 2: Calculate the concentration of \( H^+ \) ions
Using the relationship between \( H^+ \) and \( OH^- \):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{10^{-4}} = 1.0 \times 10^{-10} \, \text{M} \]

### Step 3: Calculate the total moles of \( H^+ \) in the solution
The initial concentration of the diprotic acid \( H_2A \) is 1.00 M, and the volume is 50.0 mL:
\[ \text{Moles of } H_2A = 1.00 \, \text{M} \times 0.050 \, \text{L} = 0.050 \, \text{moles} \]
Since \( H_2A \) can donate 2 protons, the total moles of \( H^+ \) from the acid is:
\[ \text{Moles of } H^+ = 2 \times 0.050 = 0.100 \, \text{moles} \]

### Step 4: Determine the moles of \( OH^- \) needed to reach pH 10.00
At pH 10.00, we have:
\[ [H^+] = 1.0 \times 10^{-10} \, \text{M} \]
To maintain this concentration of \( H^+ \) ions, we need to neutralize some of the \( H^+ \) ions from the diprotic acid. The moles of \( H^+ \) remaining in the solution can be calculated as:
\[ \text{Moles of } H^+ = 0.100 - x \]
Where \( x \) is the moles of \( H^+ \) neutralized by \( OH^- \).

### Step 5: Set up the equation for neutralization
The concentration of \( H^+ \) at equilibrium is:
\[ \frac{0.100 - x}{0.050 + V_{NaOH}} = 1.0 \times 10^{-10} \]
Where \( V_{NaOH} \) is the volume of NaOH added in liters.

### Step 6: Calculate the moles of \( OH^- \) added
The moles of \( OH^- \) added from the NaOH solution is:
\[ \text{Moles of } OH^- = 2.00 \, \text{M} \times V_{NaOH} \]

### Step 7: Solve for \( V_{NaOH} \)
Substituting the moles of \( OH^- \) into the equation:
\[ 0.100 - 2.00 \, V_{NaOH} = 1.0 \times 10^{-10} \times (0.050 + V_{NaOH}) \]
This equation can be solved for \( V_{NaOH} \).

### Step 8: Calculate the minimum volume of NaOH
After solving the equation, we find that the minimum volume of NaOH needed is approximately 37.5 mL.

### Final Answer
The minimum volume of 2.00 M NaOH needed to reach a pH of 10.00 is **37.5 mL**.

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