The pH of the solution: 5ml of `(M)/(5)`, HCl + 10 ml of `(M)/(10)` NaOH is

Options

A. 5
B. 3
C. 7
D. 8

Explanation

To find the pH of the solution formed by mixing 5 mL of \( \frac{1}{5} \) M HCl and 10 mL of \( \frac{1}{10} \) M NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl
First, we need to calculate the number of moles of HCl in the solution.

\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = \left( \frac{1}{5} \, \text{M} \right) \times (5 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}) = \frac{1}{5} \times \frac{5}{1000} = \frac{1}{1000} \, \text{mol} \]

### Step 2: Calculate the moles of NaOH
Next, we calculate the number of moles of NaOH in the solution.

\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = \left( \frac{1}{10} \, \text{M} \right) \times (10 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}) = \frac{1}{10} \times \frac{10}{1000} = \frac{1}{100} \, \text{mol} \]

### Step 3: Determine the limiting reactant
Now, we compare the moles of HCl and NaOH to determine the limiting reactant.

- Moles of HCl = \( \frac{1}{1000} \, \text{mol} \)
- Moles of NaOH = \( \frac{1}{100} \, \text{mol} \)

Since \( \frac{1}{1000} < \frac{1}{100} \), HCl is the limiting reactant.

### Step 4: Calculate the moles of excess NaOH
After the reaction, we need to find out how much NaOH is left unreacted.

\[ \text{Moles of NaOH remaining} = \text{Initial moles of NaOH} - \text{Moles of HCl} \]
\[ = \frac{1}{100} - \frac{1}{1000} = \frac{10}{1000} - \frac{1}{1000} = \frac{9}{1000} \, \text{mol} \]

### Step 5: Calculate the total volume of the solution
The total volume of the solution after mixing is:

\[ \text{Total Volume} = 5 \, \text{mL} + 10 \, \text{mL} = 15 \, \text{mL} = 0.015 \, \text{L} \]

### Step 6: Calculate the concentration of the excess NaOH
Now, we can calculate the concentration of the remaining NaOH in the total volume.

\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH remaining}}{\text{Total Volume}} = \frac{\frac{9}{1000}}{0.015} = \frac{9}{1000} \times \frac{1000}{15} = \frac{9}{15} = 0.6 \, \text{M} \]

### Step 7: Calculate the pOH and then the pH
Now, we can calculate the pOH of the solution.

\[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.6) \]

Using a calculator, we find:

\[ \text{pOH} \approx 0.22 \]

Now, we can find the pH using the relation \( \text{pH} + \text{pOH} = 14 \).

\[ \text{pH} = 14 - \text{pOH} \approx 14 - 0.22 = 13.78 \]

### Final Answer:
The pH of the solution is approximately **13.78**.

---