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].
### 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].
