20 mL of 0.1 M weak acid `HA(K_(a)=10^(-5))` is mixed with solution of 10 mL of 0.3 M HCl and 10 mL. of 0.1 M NaOH. Find the value of `[A^(-)]`//([HA]+[A^(-)])` in the resulting solution :

Options

A. `2xx10^(-4)`
B. `2xx10^(-5)`
C. `2xx10^(-3)`
D. `0.05`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the concentration of H⁺ ions
First, we need to find the total concentration of H⁺ ions in the solution after mixing the weak acid (HA), HCl, and NaOH.

1. **Calculate moles of HCl:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.3 \, \text{M} \times 0.010 \, \text{L} = 0.003 \, \text{moles} \]

2. **Calculate moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.010 \, \text{L} = 0.001 \, \text{moles} \]

3. **Determine the net moles of H⁺:**
Since HCl is a strong acid and NaOH is a strong base, they will react with each other:
\[ \text{Net moles of H⁺} = \text{Moles of HCl} - \text{Moles of NaOH} = 0.003 - 0.001 = 0.002 \, \text{moles} \]

4. **Calculate the total volume of the solution:**
\[ \text{Total Volume} = 20 \, \text{mL} + 10 \, \text{mL} + 10 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L} \]

5. **Calculate the concentration of H⁺:**
\[ [H^+] = \frac{\text{Net moles of H⁺}}{\text{Total Volume}} = \frac{0.002}{0.040} = 0.05 \, \text{M} \]

### Step 2: Calculate the concentration of HA
Next, we will calculate the concentration of the weak acid HA in the resulting solution.

1. **Calculate moles of HA:**
\[ \text{Moles of HA} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{moles} \]

2. **Calculate the concentration of HA in the total volume:**
\[ [HA] = \frac{\text{Moles of HA}}{\text{Total Volume}} = \frac{0.002}{0.040} = 0.05 \, \text{M} \]

### Step 3: Set up the equilibrium expression for HA dissociation
The dissociation of the weak acid HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

At equilibrium, we can express the concentrations as follows:
- \([HA] = 0.05 - x\)
- \([H^+] = 0.05 + x\)
- \([A^-] = x\)

### Step 4: Use the Ka expression
Using the given \(K_a = 10^{-5}\):
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(0.05 + x)(x)}{0.05 - x} \]

Assuming \(x\) is very small compared to 0.05, we can simplify:
\[ K_a \approx \frac{(0.05)(x)}{0.05} = x \]
Thus, \(x = K_a = 10^{-5}\).

### Step 5: Calculate the ratio \(\frac{[A^-]}{[HA] + [A^-]}\)
Now we can find the concentrations:
- \([A^-] = x = 10^{-5}\)
- \([HA] = 0.05\)

Now, we can calculate the ratio:
\[ \frac{[A^-]}{[HA] + [A^-]} = \frac{10^{-5}}{0.05 + 10^{-5}} \approx \frac{10^{-5}}{0.05} = \frac{10^{-5}}{5 \times 10^{-2}} = 2 \times 10^{-4} \]

### Final Answer
The value of \(\frac{[A^-]}{[HA] + [A^-]} = 2 \times 10^{-4}\).

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