200 mL of 1 HCl, is mixed with 300 mL of 6 M and the final solution is diluted to 1000 mL.calculate molar concentration of `[H^(+)`] ion .

Answer

To solve the problem step by step, we will calculate the molar concentration of the \( [H^+] \) ions after mixing the two solutions and diluting them to a final volume of 1000 mL.

### Step 1: Calculate the number of moles of HCl in each solution

1. **For the first solution (200 mL of 1 M HCl)**:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.200 \, \text{L} = 0.200 \, \text{mol} \]

2. **For the second solution (300 mL of 6 M HCl)**:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 6 \, \text{mol/L} \times 0.300 \, \text{L} = 1.800 \, \text{mol} \]

### Step 2: Calculate the total number of moles of HCl

Now, we add the moles of HCl from both solutions:
\[ \text{Total moles of HCl} = 0.200 \, \text{mol} + 1.800 \, \text{mol} = 2.000 \, \text{mol} \]

### Step 3: Calculate the final volume of the solution

The final volume after dilution is given as 1000 mL, which is equivalent to:
\[ \text{Final Volume} = 1.000 \, \text{L} \]

### Step 4: Calculate the molar concentration of \( [H^+] \)

Since HCl is a strong acid, it dissociates completely in solution, meaning that the concentration of \( [H^+] \) will be equal to the concentration of HCl. We can calculate the molarity of \( [H^+] \) using the formula:
\[ \text{Molarity} = \frac{\text{Total moles of HCl}}{\text{Final Volume in L}} = \frac{2.000 \, \text{mol}}{1.000 \, \text{L}} = 2.000 \, \text{M} \]

### Final Answer:
The molar concentration of \( [H^+] \) ions in the final solution is **2 M**.

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