10 ml of 0.1 M HCl is titrated with 0.1 M NaOH. When the volume of NaOH added from the burette is from 9.99 mL to 10.01 ml, the pH jumps approximately from

Options

A. 4 to 10
B. 6 to 8
C. 6.9 to 7.1
D. 1 to 14

Answer

To solve the problem of how the pH changes when titrating 10 mL of 0.1 M HCl with 0.1 M NaOH, we will follow these steps:

### Step 1: Understand the Initial Conditions
We start with 10 mL of 0.1 M HCl. The concentration of HCl is 0.1 M, which means that the number of moles of HCl in 10 mL is:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.01 \, \text{L} = 0.001 \, \text{mol} \]

### Step 2: Determine the Situation at 9.99 mL of NaOH
When we add 9.99 mL of 0.1 M NaOH, we need to calculate the moles of NaOH added:
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.00999 \, \text{L} = 0.000999 \, \text{mol} \]

Now, we can find the moles of HCl remaining after the reaction:
\[ \text{Moles of HCl remaining} = 0.001 \, \text{mol} - 0.000999 \, \text{mol} = 0.000001 \, \text{mol} \]

### Step 3: Calculate the Concentration of H⁺ Ions
The total volume of the solution after adding 9.99 mL of NaOH is:
\[ \text{Total Volume} = 10 \, \text{mL} + 9.99 \, \text{mL} = 19.99 \, \text{mL} = 0.01999 \, \text{L} \]
The concentration of H⁺ ions is:
\[ \text{[H⁺]} = \frac{\text{Moles of H⁺}}{\text{Total Volume}} = \frac{0.000001 \, \text{mol}}{0.01999 \, \text{L}} \approx 5.005 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Calculate the pH at 9.99 mL of NaOH
Using the concentration of H⁺ ions, we can calculate the pH:
\[ \text{pH} = -\log[H⁺] \approx -\log(5.005 \times 10^{-5}) \approx 4.30 \]

### Step 5: Determine the Situation at 10.01 mL of NaOH
When we add 10.01 mL of NaOH, we calculate the moles of NaOH added:
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.01001 \, \text{L} = 0.001001 \, \text{mol} \]

Now, we find the moles of NaOH in excess:
\[ \text{Excess NaOH} = 0.001001 \, \text{mol} - 0.001 \, \text{mol} = 0.000001 \, \text{mol} \]

### Step 6: Calculate the Concentration of OH⁻ Ions
The total volume of the solution after adding 10.01 mL of NaOH is:
\[ \text{Total Volume} = 10 \, \text{mL} + 10.01 \, \text{mL} = 20.01 \, \text{mL} = 0.02001 \, \text{L} \]
The concentration of OH⁻ ions is:
\[ \text{[OH⁻]} = \frac{0.000001 \, \text{mol}}{0.02001 \, \text{L}} \approx 5.00 \times 10^{-5} \, \text{mol/L} \]

### Step 7: Calculate the pOH and then pH at 10.01 mL of NaOH
Now we can calculate pOH:
\[ \text{pOH} = -\log[OH⁻] \approx -\log(5.00 \times 10^{-5}) \approx 4.30 \]
Finally, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} \approx 14 - 4.30 = 9.70 \]

### Conclusion
When the volume of NaOH added changes from 9.99 mL to 10.01 mL, the pH jumps approximately from **4.30 to 9.70**.

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