20 mL of 0.1 M `H_2SO_4` solution is added to 30 mL of 0.2 M `NH_4OH` solution. The pH of the resultant mixture is : [`pK_b` of `NH_4OH`=4.7]

Options

A. 9.4
B. `5.0`
C. `9.0`
D. `5.2`

Answer

To solve the problem of finding the pH of the resultant mixture when 20 mL of 0.1 M `H2SO4` is added to 30 mL of 0.2 M `NH4OH`, we can follow these steps:

### Step 1: Calculate the moles of `H2SO4` and `NH4OH`

1. **Moles of `H2SO4`:**
- Volume = 20 mL = 0.020 L
- Concentration = 0.1 M
- Moles = Concentration × Volume = 0.1 mol/L × 0.020 L = 0.002 moles of `H2SO4`
- Since `H2SO4` dissociates to give 2 moles of `H+`, the moles of `H+` produced = 2 × 0.002 = 0.004 moles.

2. **Moles of `NH4OH`:**
- Volume = 30 mL = 0.030 L
- Concentration = 0.2 M
- Moles = Concentration × Volume = 0.2 mol/L × 0.030 L = 0.006 moles of `NH4OH`.

### Step 2: Determine the reaction between `H2SO4` and `NH4OH`

- The reaction can be represented as:
\[ H2SO4 + 2 NH4OH \rightarrow (NH4)2SO4 + 2 H2O \]

- From the stoichiometry of the reaction, 1 mole of `H2SO4` reacts with 2 moles of `NH4OH`.
- We have 0.004 moles of `H2SO4` which will react with 0.008 moles of `NH4OH`. However, we only have 0.006 moles of `NH4OH`, which means `NH4OH` is the limiting reagent.

### Step 3: Calculate the remaining moles after the reaction

- Since 0.006 moles of `NH4OH` will react with 0.003 moles of `H2SO4` (because 1 mole of `H2SO4` reacts with 2 moles of `NH4OH`):
- Remaining `H2SO4` = 0.004 moles - 0.003 moles = 0.001 moles.
- `NH4OH` will be completely consumed.

### Step 4: Calculate the concentration of remaining `H2SO4`

- Total volume of the solution = 20 mL + 30 mL = 50 mL = 0.050 L.
- Concentration of remaining `H2SO4` = \(\frac{0.001 \text{ moles}}{0.050 \text{ L}} = 0.02 \text{ M}\).

### Step 5: Calculate the pH of the solution

- Since we have a strong acid (`H2SO4`), we can directly calculate the pH from the concentration of `H+` ions.
- Concentration of `H+` ions = 0.02 M.
- pH = -log[H+] = -log(0.02) ≈ 1.70.

### Final Answer:
The pH of the resultant mixture is approximately **1.70**.

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