VIPSolutionsA buffer solution contains 0 . 6 0 0 M lactic acid ( HC 3 H 5 O 3 ) and 0 . 6 2 0 M the lactate ion ( C 3 H 5 O 3 - ) . What is the resulting pH if 1 0 0 . 0 mL of 0 . 1 0 0 M potassium hydroxide is added to 0 . 5 0 0 L of the buffer solution? ( Ka of HC 3 H 5 O 3 is 1 . 4 × 1 0 - 4. Do step by step proper solution with proper explaination
Question:
A buffer solution contains 0 . 6 0 0 M lactic acid ( HC 3 H 5 O 3 ) and 0 . 6 2 0 M the lactate ion ( C 3 H 5 O 3 - ) . What is the resulting pH if 1 0 0 . 0 mL of 0 . 1 0 0 M potassium hydroxide is added to 0 . 5 0 0 L of the buffer solution? ( Ka of HC 3 H 5 O 3 is 1 . 4 × 1 0 - 4. Do step by step proper solution with proper explaination
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Asked by: Test User 100142
Created at: 2025-05-05 10:03:24
Asked by: Test User 100142
Created at: 2025-05-05 10:03:24
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Let's walk through the problem **step by step**, explaining each part.
---
## **Step 1: Write the Buffer Reaction**
Lactic acid is a weak acid, and its conjugate base is the lactate ion:
\[
\mathrm{HC_3H_5O_3 \rightleftharpoons H^+ + C_3H_5O_3^-}
\]
We are told:
- \([\mathrm{HC_3H_5O_3}] = 0.600~\mathrm{M}\)
- \([\mathrm{C_3H_5O_3^-}] = 0.620~\mathrm{M}\)
- Volume of buffer = \(0.500~\mathrm{L}\)
- \(K_a = 1.4 \times 10^{-4}\)
---
## **Step 2: Moles of Acid and Base in the Buffer**
Calculate the **moles** of each species in the buffer:
\[
\text{Moles of lactic acid (HA):}~0.600~\mathrm{mol/L} \times 0.500~\mathrm{L} = 0.300~\mathrm{mol}
\]
\[
\text{Moles of lactate ion (A⁻):}~0.620~\mathrm{mol/L} \times 0.500~\mathrm{L} = 0.310~\mathrm{mol}
\]
---
## **Step 3: Moles of KOH Added**
\[
\text{Volume of KOH added} = 100.0~\mathrm{mL} = 0.100~\mathrm{L}
\]
\[
\text{Concentration of KOH} = 0.100~\mathrm{M}
\]
\[
\text{Moles of KOH added} = 0.100~\mathrm{mol/L} \times 0.100~\mathrm{L} = 0.0100~\mathrm{mol}
\]
---
## **Step 4: Buffer Reaction with KOH**
KOH is a strong base. It reacts completely with lactic acid:
\[
\mathrm{HC_3H_5O_3} + \mathrm{OH^-} \rightarrow \mathrm{C_3H_5O_3^-} + \mathrm{H_2O}
\]
So, for every 1 mole of KOH, 1 mole of lactic acid is neutralized and 1 mole of lactate ion is produced.
- **Moles of lactic acid after reaction:** \(0.300~\mathrm{mol} - 0.0100~\mathrm{mol} = 0.290~\mathrm{mol}\)
- **Moles of lactate after reaction:** \(0.310~\mathrm{mol} + 0.0100~\mathrm{mol} = 0.320~\mathrm{mol}\)
---
## **Step 5: New Concentrations (After Addition, New Volume)**
The **total volume** is now:
\[
0.500~\mathrm{L} + 0.100~\mathrm{L} = 0.600~\mathrm{L}
\]
So:
\[
[\mathrm{HC_3H_5O_3}] = \frac{0.290~\mathrm{mol}}{0.600~\mathrm{L}} = 0.483~\mathrm{M}
\]
\[
[\mathrm{C_3H_5O_3^-}] = \frac{0.320~\mathrm{mol}}{0.600~\mathrm{L}} = 0.533~\mathrm{M}
\]
---
## **Step 6: Calculate pH Using Henderson-Hasselbalch Equation**
\[
\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right)
\]
Where:
\[
\mathrm{p}K_a = -\log(1.4 \times 10^{-4}) = 3.85
\]
Now:
\[
\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{0.533}{0.483} = 1.103
\]
\[
\log(1.103) \approx 0.043
\]
So:
\[
\mathrm{pH} = 3.85 + 0.043 = 3.89
\]
---
## **Final Answer**
\[
\boxed{\text{The resulting pH after adding the KOH is } \mathbf{3.89}}
\]
---
### **Summary Table**
| Species | Initial moles | Change (KOH added) | Final moles | Final concentration (0.600 L) |
|------------------------|---------------|--------------------|-------------|-------------------------------|
| Lactic acid (HA) | 0.300 | \(-0.0100\) | 0.290 | 0.483 M |
| Lactate ion (A⁻) | 0.310 | \(+0.0100\) | 0.320 | 0.533 M |
---
### **Key Steps:**
1. Find moles of acid and base before and after reaction.
2. Update concentrations for new total volume.
3. Use the Henderson-Hasselbalch equation to find the new pH.
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