`pH` calculation upon dilute of strong acid solution is generally done by equating `n_(H)` in original solution & diluted solution.However . If strong acid solution is very dilute then `H^(+)` from water are also to be considered take `log3.7=0.568` and answer the following questions.
A `1` litres solution of `pH=4`(solution of a strong acid ) is added to the `7//3` litres of water.What is the `pH `of resulting solution?

Options

A. `4.52`
B. `4.365`
C. `4.4`
D. `4.432`

Explanation

To solve the problem of calculating the pH of the resulting solution after diluting a strong acid solution, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions in the original solution
Given that the pH of the original solution is 4, we can calculate the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]
Substituting the value of pH:
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