2. If you prepared a solution of KOH by weighing out 6.255 g of solid KOH that had a purity of 87.5%
and then diluted this to a final volume of 1 L in a volumetric flask, what would be the %KHP in
an unknown sample (mass = 0.601 g) if it required 21.36 mL of KOH titrant to reach the endpoint?
Show calculations.



Answer :

Sure, let's solve this step-by-step:

### Step 1: Calculate the moles of pure KOH in the solution

We are given:

- Mass of solid KOH = 6.255 g
- Purity of KOH = 87.5%

First, convert the purity percentage to a fraction:

[tex]\[ \text{Purity of KOH} = \frac{87.5}{100} = 0.875 \][/tex]

Now, calculate the mass of pure KOH:

[tex]\[ \text{Mass of pure KOH} = \text{Mass of solid KOH} \times \text{Purity} \][/tex]
[tex]\[ \text{Mass of pure KOH} = 6.255 \, \text{g} \times 0.875 \][/tex]
[tex]\[ \text{Mass of pure KOH} = 5.469 \, \text{g} \][/tex]

Next, use the molar mass of KOH to find the moles of KOH. The molar mass of KOH is 56.11 g/mol.

[tex]\[ \text{Moles of KOH} = \frac{\text{Mass of pure KOH}}{\text{Molar mass of KOH}} \][/tex]
[tex]\[ \text{Moles of KOH} = \frac{5.469 \, \text{g}}{56.11 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of KOH} = 0.0975 \, \text{mol} \][/tex]

### Step 2: Calculate the concentration of the KOH solution

The KOH solution was diluted to a final volume of 1 L:

[tex]\[ \text{Concentration of KOH} = \frac{\text{Moles of KOH}}{\text{Volume of solution}} \][/tex]
[tex]\[ \text{Concentration of KOH} = \frac{0.0975 \, \text{mol}}{1 \, \text{L}} \][/tex]
[tex]\[ \text{Concentration of KOH} = 0.0975 \, \text{mol/L} \][/tex]

### Step 3: Calculate the moles of KOH used in the titration

Given that 21.36 mL of KOH titrant was used:

[tex]\[ \text{Volume of KOH titrant} = 21.36 \, \text{mL} = 0.02136 \, \text{L} \][/tex]

Now calculate the moles of KOH used:

[tex]\[ \text{Moles of KOH used} = \text{Concentration of KOH} \times \text{Volume of KOH titrant} \][/tex]
[tex]\[ \text{Moles of KOH used} = 0.0975 \, \text{mol/L} \times 0.02136 \, \text{L} \][/tex]
[tex]\[ \text{Moles of KOH used} = 0.002082 \, \text{mol} \][/tex]

### Step 4: Calculate the moles of KHP that reacted with KOH

The titration reaction between KOH and KHP (Potassium Hydrogen Phthalate) is:

[tex]\[ \text{KOH} + \text{KHP} \rightarrow \text{K}^+ + \text{HP}^- + \text{H}_2\text{O} \][/tex]

Since the mole ratio between KOH and KHP is 1:1, the moles of KOH used will be equal to the moles of KHP that reacted:

[tex]\[ \text{Moles of KHP} = 0.002082 \, \text{mol} \][/tex]

### Step 5: Calculate the mass of KHP that reacted

The molar mass of KHP is 204.22 g/mol:

[tex]\[ \text{Mass of KHP} = \text{Moles of KHP} \times \text{Molar mass of KHP} \][/tex]
[tex]\[ \text{Mass of KHP} = 0.002082 \, \text{mol} \times 204.22 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of KHP} = 0.4251 \, \text{g} \][/tex]

### Step 6: Calculate the percentage of KHP in the unknown sample

The total mass of the unknown sample is given as 0.601 g. Now, calculate the percentage of KHP in this sample:

[tex]\[ \text{Percentage of KHP} = \left( \frac{\text{Mass of KHP}}{\text{Mass of sample}} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage of KHP} = \left( \frac{0.4251 \, \text{g}}{0.601 \, \text{g}} \right) \times 100 \][/tex]
[tex]\[ \text{Percentage of KHP} \approx 70.74\% \][/tex]

### Final Answer

The percentage of KHP in the unknown sample is approximately [tex]\( 70.74\% \)[/tex].