Find `DeltapH` when 100 ml of 0.01 M HCl is added in a solution containing 0.1 m moles of `NaHCO_(3)` solution of negligible volume. `(Ka_(1)=10^(-7), Ka_(2)=10^(-11)` for `H_(2)CO_(3))`

Explanation

To solve the problem of finding the change in pH (ΔpH) when 100 ml of 0.01 M HCl is added to a solution containing 0.1 millimoles of NaHCO₃, we can follow these steps:

### Step 1: Calculate the initial pH of the NaHCO₃ solution
NaHCO₃ is a weak base, and its pH can be calculated using the formula:
\[ \text{pH} = 14 - \text{pKa} \]
Given that \( K_a1 \) for \( H_2CO_3 \) is \( 10^{-7} \), we can find \( pK_a1 \):
\[ pK_a1 = -\log(10^{-7}) = 7 \]
Thus, the pH of the NaHCO₃ solution is:
\[ \text{pH} = 14 - 7 = 7 \]

### Step 2: Calculate the moles of HCl added
The volume of HCl solution is 100 ml (0.1 L) with a concentration of 0.01 M:
\[ \text{moles of HCl} = \text{Volume} \times \text{Concentration} = 0.1 \, \text{L} \times 0.01 \, \text{mol/L} = 0.001 \, \text{mol} = 1 \, \text{mmol} \]

### Step 3: Determine the reaction between HCl and NaHCO₃
The reaction between HCl and NaHCO₃ can be represented as:
\[ \text{H}^+ + \text{HCO}_3^- \rightarrow \text{H}_2CO_3 \]
Initially, we have 0.1 mmol of NaHCO₃ and 1 mmol of HCl. The HCl will react with NaHCO₃, consuming it.

### Step 4: Calculate the remaining moles of NaHCO₃ after reaction
Since we have 1 mmol of HCl and 0.1 mmol of NaHCO₃:
\[ \text{Remaining NaHCO}_3 = 0.1 \, \text{mmol} - 1 \, \text{mmol} = 0 \, \text{mmol} \, \text{(all NaHCO₃ is consumed)} \]
Thus, all NaHCO₃ is converted to \( H_2CO_3 \).

### Step 5: Calculate the final pH after the reaction
Since all NaHCO₃ is converted to \( H_2CO_3 \), we can use the \( K_a \) values to find the pH. The concentration of \( H_2CO_3 \) formed is:
\[ \text{Concentration of } H_2CO_3 = \frac{0.1 \, \text{mmol}}{0.1 \, \text{L}} = 0.01 \, \text{M} \]
Using the \( K_a \) for \( H_2CO_3 \):
\[ K_a = 10^{-7} \]
Using the formula for \( K_a \):
\[ K_a = \frac{[H^+][HCO_3^-]}{[H_2CO_3]} \]
Assuming \( x \) is the concentration of \( H^+ \) produced:
\[ 10^{-7} = \frac{x^2}{0.01 - x} \approx \frac{x^2}{0.01} \]
Solving for \( x \):
\[ x^2 = 10^{-7} \times 0.01 = 10^{-9} \]
\[ x = 10^{-4.5} \approx 3.16 \times 10^{-5} \, \text{M} \]
Thus, the final pH is:
\[ \text{pH} = -\log(3.16 \times 10^{-5}) \approx 4.5 \]

### Step 6: Calculate ΔpH
Now, we can find the change in pH:
\[ \Delta pH = \text{Final pH} - \text{Initial pH} = 4.5 - 7 = -2.5 \]

### Final Answer
\[ \Delta pH = -2.5 \]