When 100 ml of N/10 NaOH are added to 50 ml of N/5 HCI, the `P^(H)` of the resulting solution is

Options

A. 7
B. Greater than 7
C. less than 7
D. zero

Answer

To find the pH of the resulting solution when 100 mL of N/10 NaOH is added to 50 mL of N/5 HCl, we can follow these steps:

### Step 1: Calculate the equivalents of NaOH
The formula to calculate equivalents is:
\[ \text{Equivalents} = \text{Normality} \times \text{Volume (in L)} \]
For NaOH:
- Normality (N) = N/10 = 0.1 N
- Volume = 100 mL = 0.1 L

Calculating the equivalents:
\[ \text{Equivalents of NaOH} = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} = 10 \, \text{milli-equivalents} \]

### Step 2: Calculate the equivalents of HCl
Using the same formula for HCl:
- Normality (N) = N/5 = 0.2 N
- Volume = 50 mL = 0.05 L

Calculating the equivalents:
\[ \text{Equivalents of HCl} = 0.2 \, \text{N} \times 0.05 \, \text{L} = 0.01 \, \text{equivalents} = 10 \, \text{milli-equivalents} \]

### Step 3: Compare the equivalents of NaOH and HCl
We found that:
- Equivalents of NaOH = 10 milli-equivalents
- Equivalents of HCl = 10 milli-equivalents

Since the equivalents of NaOH and HCl are equal, they completely neutralize each other.

### Step 4: Determine the pH of the resulting solution
When a strong acid (HCl) is completely neutralized by a strong base (NaOH), the resulting solution is neutral. Therefore, the pH of the resulting solution will be:
\[ \text{pH} = 7 \]

### Final Answer:
The pH of the resulting solution is **7**.

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