A 1.025 g sample containing a weak acid HX (mol. Mass=82) is dissolved in 60 mL. water and titrated with 0.25 M NaOH. When half of the acid was neutralised the pH was found to be 5.0 and at the equivalence point the pH is 9.0. Calculate mass precentage of HX in sample :
a.`50%`
b.`75%`
c.`80%`
d.None

Options

A. `50%`
B. `75%`
C. `80%`
D. None of these

Answer

To solve the problem step-by-step, we will follow the information provided in the question and the video transcript.

### Step 1: Determine the moles of weak acid HX in the sample
Given:
- Mass of the sample = 1.025 g
- Molar mass of HX = 82 g/mol

To find the number of moles of HX:
\[ \text{Moles of HX} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.025 \, \text{g}}{82 \, \text{g/mol}} \approx 0.0125 \, \text{mol} \]

### Step 2: Use the pH at half-neutralization to find pKa
At half-neutralization, the pH is equal to the pKa of the weak acid. Given:
- pH at half-neutralization = 5.0

Thus, we have:
\[ \text{pKa} = 5.0 \]

### Step 3: Determine the pH at the equivalence point
Given:
- pH at the equivalence point = 9.0

Using the relationship at the equivalence point:
\[ \text{pH} = \frac{1}{2} \text{pK}_w + \text{pKa} + \log C \]
Where \( \text{pK}_w = 14 \). Therefore:
\[ 9.0 = \frac{1}{2} \times 14 + 5 + \log C \]
\[ 9.0 = 7 + 5 + \log C \]
\[ 9.0 = 12 + \log C \]
\[ \log C = 9.0 - 12 = -3 \]
Thus:
\[ C = 10^{-3} = 0.001 \, \text{mol/L} \]

### Step 4: Relate the concentration to the volume of NaOH used
Let \( V \) be the volume of NaOH used in liters. The moles of NaOH used can be expressed as:
\[ \text{Moles of NaOH} = C \times V = 0.001 \times V \]

Since at equivalence point, moles of NaOH = moles of HX:
\[ 0.001 \times V = 0.0125 \]
\[ V = \frac{0.0125}{0.001} = 12.5 \, \text{L} \text{ or } 12.5 \, \text{mL} \]

### Step 5: Calculate the mass of HX neutralized
The moles of HX neutralized can be calculated as:
\[ \text{Moles of HX neutralized} = 0.25 \times V \]
Where \( V \) is the volume of NaOH used (in mL). Thus:
\[ \text{Moles of HX neutralized} = 0.25 \times 12.5 = 3.125 \, \text{mmol} = 0.003125 \, \text{mol} \]

### Step 6: Calculate the mass of HX
The mass of HX can be calculated as:
\[ \text{Mass of HX} = \text{moles} \times \text{molar mass} = 0.003125 \, \text{mol} \times 82 \, \text{g/mol} \approx 0.25625 \, \text{g} \]

### Step 7: Calculate the mass percentage of HX in the sample
The mass percentage of HX in the sample is given by:
\[ \text{Mass percentage of HX} = \left( \frac{\text{mass of HX}}{\text{mass of sample}} \right) \times 100 \]
\[ \text{Mass percentage of HX} = \left( \frac{0.25625}{1.025} \right) \times 100 \approx 25.0\% \]

### Conclusion
The mass percentage of HX in the sample is approximately 25.0%. Since this value does not match any of the options provided (50%, 75%, 80%, None), the correct answer is **d. None**.