The `pH` of `0.1 M` `K_(2)SO_(4)` is `7`. The solutin is diluted by `10` times. Then the pH of resulting solution is

Options

A. `7`
B. `6`
C. `8`
D. `14`

Answer

To solve the problem step by step, we need to analyze the situation involving the potassium sulfate (K₂SO₄) solution and its pH after dilution.

### Step 1: Understand the Initial Conditions
- We have a solution of K₂SO₄ with a concentration of 0.1 M.
- The pH of this solution is given as 7.

### Step 2: Recognize the Nature of K₂SO₄
- K₂SO₄ is a salt formed from a strong acid (H₂SO₄) and a strong base (KOH).
- Since it is derived from a strong acid and a strong base, K₂SO₄ is a neutral salt.

### Step 3: Consider the Effect of Dilution
- The solution is diluted by a factor of 10. This means the new concentration of K₂SO₄ will be:
\[ \text{New concentration} = \frac{0.1 \, \text{M}}{10} = 0.01 \, \text{M} \]
- However, since K₂SO₄ is a neutral salt, the dilution will not affect the pH.

### Step 4: Determine the pH After Dilution
- The pH of a neutral solution remains 7 regardless of the concentration of the neutral salt, as long as no additional acids or bases are introduced.
- Therefore, after dilution, the pH of the resulting solution will still be 7.

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

---