`7.5 gm` glycine is added to `500 ml` water. What is the `pH` of the
solution ? (Given: `pK_(a_(1)) ("glycine") = 2.4, pK_(a_(2)) ("glycine") = 9.54` )

Options

A. `6.15`
B. `2.58`
C. `9.58`
D. `5.97`

Explanation

To find the pH of the solution when 7.5 g of glycine is added to 500 mL of water, we can follow these steps:

### Step 1: Determine the Molar Mass of Glycine
Glycine (NH₂CH₂COOH) has a molar mass of approximately 75 g/mol.

### Step 2: Calculate the Number of Moles of Glycine
To find the number of moles of glycine in 7.5 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles} = \frac{7.5 \, \text{g}}{75 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Calculate the Concentration of Glycine
Since the glycine is dissolved in 500 mL of water, we can calculate the concentration (C) in mol/L:
\[ C = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
\[ C = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{mol/L} \]

### Step 4: Determine the pKa Values
We are given:
- \( pK_a1 = 2.4 \)
- \( pK_a2 = 9.54 \)

### Step 5: Calculate the Isoelectric Point (pH)
The pH at the isoelectric point (pI) for glycine can be calculated using the formula:
\[ pI = \frac{pK_a1 + pK_a2}{2} \]
Substituting the values:
\[ pI = \frac{2.4 + 9.54}{2} = \frac{11.94}{2} = 5.97 \]

### Step 6: Conclusion
The pH of the solution when 7.5 g of glycine is added to 500 mL of water is approximately **5.97**.

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