An aqueous soluton contain a substance which yields ` 4 xx 10^(-3)` mol `"litre"^(-1)` ion of `H_(3)O^(+)` . If log 2 is 0.3010 the pH of the solution is :

Options

A. 2.398
B. 1.5
C. 3
D. 3.4

Explanation

To find the pH of the solution given that it yields \( 4 \times 10^{-3} \) mol/L of \( H_3O^+ \), we can follow these steps:

### Step 1: Understand the relationship between \( H_3O^+ \) and pH
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
In aqueous solutions, \( H^+ \) ions are often represented as \( H_3O^+ \) ions. Therefore, we can write:
\[ \text{pH} = -\log[H_3O^+] \]

### Step 2: Substitute the concentration of \( H_3O^+ \)
We know from the problem that the concentration of \( H_3O^+ \) is \( 4 \times 10^{-3} \) mol/L. Thus, we can substitute this value into the pH formula:
\[ \text{pH} = -\log(4 \times 10^{-3}) \]

### Step 3: Use logarithmic properties
Using the property of logarithms that states \( \log(ab) = \log a + \log b \), we can break this down:
\[ \text{pH} = -(\log 4 + \log(10^{-3})) \]
We know that \( \log(10^{-3}) = -3 \), so we can rewrite this as:
\[ \text{pH} = -(\log 4 - 3) \]
This simplifies to:
\[ \text{pH} = 3 - \log 4 \]

### Step 4: Express \( \log 4 \) in terms of \( \log 2 \)
Since \( \log 4 = \log(2^2) = 2 \log 2 \), we can substitute this into our equation:
\[ \text{pH} = 3 - 2 \log 2 \]

### Step 5: Substitute the value of \( \log 2 \)
Given that \( \log 2 = 0.3010 \), we can now substitute this value:
\[ \text{pH} = 3 - 2 \times 0.3010 \]
Calculating this gives:
\[ \text{pH} = 3 - 0.602 = 2.398 \]

### Final Answer
Thus, the pH of the solution is:
\[ \text{pH} = 2.398 \]