Find the `DeltapH(` initial `pH -` final `pH)` when `100 ml 0.01M HCl` is added in a solution containing `0.1m` moles of `NaHCO_(3)` solution of negligible volume initial` pH =9,k_(a_(1))=10^(-7),k_(a_(2))=10^(-11)` for `H_(2)CO_(3)) :`

Options

A. `6+2 log 3`
B. `3-2 log 3`
C. `3+2log2`
D. `6-2log 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 moles of NaHCO₃, we can follow these steps:

### Step 1: Calculate Initial pH
The initial pH of the solution is given as 9.

### Step 2: Calculate Moles of HCl
To find the moles of HCl in the solution, we use the formula:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (L)} \]
Given that the molarity of HCl is 0.01 M and the volume is 100 mL (0.1 L):
\[ \text{Moles of HCl} = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} = 0.001 \, \text{moles} \]

### Step 3: Calculate Moles of NaHCO₃
The moles of NaHCO₃ are given as 0.1 moles.

### Step 4: Determine the Limiting Reagent
The reaction between HCl and NaHCO₃ can be written as:
\[ \text{HCl} + \text{NaHCO}_3 \rightarrow \text{NaCl} + \text{H}_2\text{CO}_3 \]
Since we have 0.001 moles of HCl and 0.1 moles of NaHCO₃, HCl is the limiting reagent.

### Step 5: Calculate Remaining Moles After Reaction
After the reaction, the moles of HCl consumed will be equal to the moles of NaHCO₃ reacted, which is 0.001 moles. Therefore:
- Moles of HCl remaining = 0.001 - 0.001 = 0 moles
- Moles of NaHCO₃ remaining = 0.1 - 0.001 = 0.099 moles

### Step 6: Calculate Final pH
Since all HCl has reacted, we need to find the concentration of H⁺ ions from the remaining NaHCO₃. The remaining NaHCO₃ will partially dissociate to form H₂CO₃ and subsequently release H⁺ ions.

The equilibrium for the dissociation of H₂CO₃ can be described by its dissociation constants:
- \( K_{a1} = 10^{-7} \) for \( H_2CO_3 \leftrightarrow H^+ + HCO_3^- \)
- \( K_{a2} = 10^{-11} \) for \( HCO_3^- \leftrightarrow H^+ + CO_3^{2-} \)

Using the first dissociation, we can find the concentration of H⁺:
Let \( x \) be the concentration of H⁺ produced from the dissociation of 0.099 moles of NaHCO₃ in a total volume of 100 mL (0.1 L):
\[ [H^+] = \frac{x}{0.1} \]

Using the equilibrium expression for the first dissociation:
\[ K_{a1} = \frac{[H^+][HCO_3^-]}{[H_2CO_3]} \]
Assuming \( [HCO_3^-] \approx 0.099 \) and \( [H_2CO_3] \approx 0 \) (as it is formed from NaHCO₃):
\[ 10^{-7} = \frac{x(0.099)}{0} \]

This indicates that we need to consider the second dissociation to find the final concentration of H⁺ ions.

### Step 7: Calculate Final pH
Assuming that the contribution of H⁺ from the second dissociation is negligible compared to the first, we can approximate:
\[ [H^+] \approx 10^{-7} \]
Thus:
\[ \text{Final pH} = -\log(10^{-7}) = 7 \]

### Step 8: Calculate ΔpH
Now we can calculate the change in pH:
\[ \Delta pH = \text{Initial pH} - \text{Final pH} = 9 - 7 = 2 \]

### Final Answer
The change in pH (ΔpH) is 2.

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