10 mL of 1 NHCl is mixed with 20 mL of 1`MH_(2)SO_(4)` and 30 mL of 1M NaOH. The resultant solution has:

Options

A. 20 meq of `H^(+)` ions
B. 20 meq of `OH^(-)`
C. 0 meq of `H^(+)` or `OH^(-)`
D. 30 milli moles of `OH^(-)`

Explanation

To solve the problem, we need to calculate the milliequivalents of the acids and bases involved in the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the milliequivalents of HCl
- **Given:**
- Volume of HCl = 10 mL
- Normality of HCl = 1 N

- **Formula:**
\[ \text{Milliequivalents of HCl} = \text{Normality} \times \text{Volume (in L)} \]

- **Calculation:**
\[ \text{Milliequivalents of HCl} = 1 \, \text{N} \times 10 \, \text{mL} = 1 \times 10 = 10 \, \text{mEq} \]

### Step 2: Calculate the milliequivalents of H₂SO₄
- **Given:**
- Volume of H₂SO₄ = 20 mL
- Molarity of H₂SO₄ = 1 M
- n-factor of H₂SO₄ = 2 (since it can donate 2 H⁺ ions)

- **Formula:**
\[ \text{Milliequivalents of H₂SO₄} = \text{Molarity} \times \text{n-factor} \times \text{Volume (in L)} \]

- **Calculation:**
\[ \text{Milliequivalents of H₂SO₄} = 1 \, \text{M} \times 2 \times 20 \, \text{mL} = 2 \times 20 = 40 \, \text{mEq} \]

### Step 3: Calculate the total milliequivalents of H⁺ ions
- **Total H⁺ from acids:**
\[ \text{Total H⁺} = \text{Milliequivalents of HCl} + \text{Milliequivalents of H₂SO₄} \]

- **Calculation:**
\[ \text{Total H⁺} = 10 \, \text{mEq} + 40 \, \text{mEq} = 50 \, \text{mEq} \]

### Step 4: Calculate the milliequivalents of NaOH
- **Given:**
- Volume of NaOH = 30 mL
- Molarity of NaOH = 1 M
- n-factor of NaOH = 1 (since it can donate 1 OH⁻ ion)

- **Formula:**
\[ \text{Milliequivalents of NaOH} = \text{Molarity} \times \text{n-factor} \times \text{Volume (in L)} \]

- **Calculation:**
\[ \text{Milliequivalents of NaOH} = 1 \, \text{M} \times 1 \times 30 \, \text{mL} = 1 \times 30 = 30 \, \text{mEq} \]

### Step 5: Determine the excess H⁺ ions
- **Excess H⁺:**
\[ \text{Excess H⁺} = \text{Total H⁺} - \text{Milliequivalents of NaOH} \]

- **Calculation:**
\[ \text{Excess H⁺} = 50 \, \text{mEq} - 30 \, \text{mEq} = 20 \, \text{mEq} \]

### Conclusion
The resultant solution has **20 milliequivalents of H⁺ ions remaining**.

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