Calculate the pH of a buffer solution prepared by dissolving 30g of `Na_(2)CO_(3)` in 500 mL of an aqueous solution contaning 150 mL of 1 M HCI. `(K_(a) for HCO_(3)^(-)= 5.63xx10^(-11))`

Answer

To calculate the pH of the buffer solution prepared by dissolving 30g of Na₂CO₃ in 500 mL of an aqueous solution containing 150 mL of 1 M HCl, we can follow these steps:

### Step 1: Calculate the moles of Na₂CO₃
First, we need to find the number of moles of Na₂CO₃ in 30g.

- Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g/mol
- Moles of Na₂CO₃ = mass (g) / molar mass (g/mol) = 30 g / 106 g/mol = 0.283 moles

### Step 2: Calculate the moles of HCl
Next, we calculate the moles of HCl in 150 mL of 1 M solution.

- Moles of HCl = concentration (mol/L) × volume (L) = 1 mol/L × 0.150 L = 0.150 moles

### Step 3: Determine the reaction between Na₂CO₃ and HCl
The reaction between Na₂CO₃ and HCl is as follows:

\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{CO}_3 \]

From the reaction, we see that 1 mole of Na₂CO₃ reacts with 2 moles of HCl.

### Step 4: Calculate the amount of Na₂CO₃ and HCl after the reaction
Since we have 0.283 moles of Na₂CO₃ and 0.150 moles of HCl, we can determine how much of each reactant remains after the reaction.

- HCl will react with Na₂CO₃:
- Moles of HCl required = 0.150 moles (which will react with 0.075 moles of Na₂CO₃)
- Remaining Na₂CO₃ = 0.283 moles - 0.075 moles = 0.208 moles
- Remaining HCl = 0.150 moles - 0.150 moles = 0 moles (all HCl is consumed)

### Step 5: Calculate the concentration of NaHCO₃ formed
Since all the HCl is consumed, Na₂CO₃ will convert to NaHCO₃. The moles of NaHCO₃ formed will be equal to the moles of HCl reacted, which is 0.075 moles.

### Step 6: Calculate the concentrations of NaHCO₃ and Na₂CO₃ in the final solution
The total volume of the solution after mixing is 500 mL + 150 mL = 650 mL = 0.650 L.

- Concentration of NaHCO₃ = moles / volume = 0.075 moles / 0.650 L = 0.115 mol/L
- Concentration of Na₂CO₃ = remaining moles / volume = 0.208 moles / 0.650 L = 0.320 mol/L

### Step 7: Use the Henderson-Hasselbalch equation to calculate pH
The Henderson-Hasselbalch equation is given by:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

Where:
- \(\text{pK}_a = -\log(K_a)\)
- \(K_a\) for HCO₃⁻ = \(5.63 \times 10^{-11}\)

Calculate pKₐ:

\[ \text{pK}_a = -\log(5.63 \times 10^{-11}) \approx 10.25 \]

Now substitute the values into the Henderson-Hasselbalch equation:

\[ \text{pH} = 10.25 + \log\left(\frac{0.320}{0.115}\right) \]

Calculate the log term:

\[ \log\left(\frac{0.320}{0.115}\right) \approx \log(2.783) \approx 0.444 \]

Now calculate pH:

\[ \text{pH} = 10.25 + 0.444 \approx 10.694 \]

### Final Answer:
The pH of the buffer solution is approximately **10.69**.

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