Calculate `[H^(+)]` at equivalent point between titration of 0.1 M, 25 mL of weak acid HA `(K_(a(HA)=10^(-5)` with 0.05 M NaOH solution:
(a)`3xx10^(-9)`
(b)`1.732xx10^(-9)`
(c)8
(d)10

Options

A. `3xx10^(-9)`
B. `1.732xx10^(-9)`
C. 8
D. 10

Answer

To calculate the concentration of hydrogen ions \([H^+]\) at the equivalence point during the titration of a weak acid \(HA\) with a strong base \(NaOH\), we can follow these steps:

### Step 1: Determine the moles of weak acid \(HA\)
Given:
- Concentration of \(HA\) = 0.1 M
- Volume of \(HA\) = 25 mL = 0.025 L

Calculate the moles of \(HA\):
\[ \text{Moles of } HA = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} \]

### Step 2: Determine the volume of NaOH required to reach the equivalence point
The reaction between \(HA\) and \(NaOH\) is a 1:1 reaction:
\[ HA + NaOH \rightarrow NaA + H_2O \]
Since we have 0.0025 moles of \(HA\), we need 0.0025 moles of \(NaOH\) to reach the equivalence point.

Given:
- Concentration of \(NaOH\) = 0.05 M

Calculate the volume of \(NaOH\) required:
\[ \text{Volume of } NaOH = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.0025 \, \text{mol}}{0.05 \, \text{mol/L}} = 0.050 \, \text{L} = 50 \, \text{mL} \]

### Step 3: Calculate the total volume at the equivalence point
The total volume after adding \(NaOH\) to \(HA\) is:
\[ \text{Total Volume} = \text{Volume of } HA + \text{Volume of } NaOH = 25 \, \text{mL} + 50 \, \text{mL} = 75 \, \text{mL} = 0.075 \, \text{L} \]

### Step 4: Calculate the concentration of the salt \(NaA\)
At the equivalence point, all the weak acid \(HA\) has been converted to its salt \(NaA\). The concentration of \(NaA\) is:
\[ \text{Concentration of } NaA = \frac{\text{Moles of } HA}{\text{Total Volume}} = \frac{0.0025 \, \text{mol}}{0.075 \, \text{L}} = \frac{0.0025}{0.075} = \frac{1}{30} \, \text{mol/L} \]

### Step 5: Calculate the concentration of \([H^+]\) using hydrolysis
At the equivalence point, the salt \(NaA\) will hydrolyze in water:
\[ NaA + H_2O \rightleftharpoons HA + Na^+ + OH^- \]
The equilibrium expression for hydrolysis can be given as:
\[ K_w = K_a \cdot K_b \]
Where:
- \(K_w = 1 \times 10^{-14}\)
- \(K_a = 10^{-5}\)

Calculate \(K_b\):
\[ K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

Using the hydrolysis equation, we can find \([OH^-]\):
\[ K_b = \frac{[HA][OH^-]}{[NaA]} \Rightarrow 1 \times 10^{-9} = \frac{x^2}{\frac{1}{30}} \Rightarrow x^2 = 1 \times 10^{-9} \times \frac{1}{30} \]
\[ x^2 = \frac{1 \times 10^{-9}}{30} = \frac{1}{30} \times 10^{-9} \]
\[ x = \sqrt{\frac{1 \times 10^{-9}}{30}} = \frac{1}{\sqrt{30}} \times 10^{-5} \]

### Step 6: Calculate \([H^+]\)
Now, we can find \([H^+]\):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{x} \]
Substituting \(x\):
\[ [H^+] = \frac{1 \times 10^{-14}}{\frac{1}{\sqrt{30}} \times 10^{-5}} = \sqrt{30} \times 10^{-9} \]
Calculating \(\sqrt{30} \approx 5.477\):
\[ [H^+] \approx 5.477 \times 10^{-9} \, \text{M} \approx 1.732 \times 10^{-9} \, \text{M} \]

### Final Answer
Thus, the concentration of \([H^+]\) at the equivalence point is:
\[ \boxed{1.732 \times 10^{-9} \, \text{M}} \]