Equal volumes of three acid solutions of `pH' 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture?

Options

A. `1.11xx10^(-4)M`
B. `3.7xx10^(-4)M`
C. `3.7xx10^(-3)M`
D. `1.11xx10^(-3)M`

Explanation

To find the concentration of \( H^+ \) ions when equal volumes of three acid solutions with pH values of 3, 4, and 5 are mixed, we can follow these steps:

### Step 1: Calculate the \( H^+ \) ion concentration for each solution.

1. **For pH 3:**
\[ [H^+] = 10^{-pH} = 10^{-3} \, \text{M} \]

2. **For pH 4:**
\[ [H^+] = 10^{-pH} = 10^{-4} \, \text{M} \]

3. **For pH 5:**
\[ [H^+] = 10^{-pH} = 10^{-5} \, \text{M} \]

### Step 2: Sum the \( H^+ \) ion concentrations.

Now, we add the concentrations of \( H^+ \) ions from all three solutions:
\[ \text{Total } [H^+] = 10^{-3} + 10^{-4} + 10^{-5} \]

### Step 3: Convert to a common format.

To add these values, we can express them in terms of the same power of ten:
\[ 10^{-3} = 1.0 \times 10^{-3} \\ 10^{-4} = 0.1 \times 10^{-3} \\ 10^{-5} = 0.01 \times 10^{-3} \]

Now adding them:
\[ \text{Total } [H^+] = 1.0 \times 10^{-3} + 0.1 \times 10^{-3} + 0.01 \times 10^{-3} = 1.11 \times 10^{-3} \, \text{M} \]

### Step 4: Calculate the concentration in the final mixture.

Since equal volumes of the three solutions are mixed, we divide the total concentration by 3 (because we have three solutions):
\[ \text{Concentration of } [H^+] = \frac{1.11 \times 10^{-3}}{3} = 0.37 \times 10^{-3} \, \text{M} = 3.7 \times 10^{-4} \, \text{M} \]

### Final Answer:
The concentration of \( H^+ \) ions in the mixture is \( 3.7 \times 10^{-4} \, \text{M} \).

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