A sample of peanut oil weighing 2g is added to 25ml of 0.4M koh. After saponification is complete, 8.5ml of 0.28M hes04 is needed to neutralize excess of koh. Saponification number of peanut oil
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The answer 146.75
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To calculate the saponification number of peanut oil, you need to use the given information and perform a few calculations. Let's break it down step-by-step:
Step 1: Calculate the amount of KOH used in the saponification reaction.
Given:
- Volume of 0.4 M KOH solution = 25 mL (or 0.025 L)
- Molarity of KOH solution = 0.4 M
To calculate the number of moles of KOH used, use the formula:
Moles KOH = Molarity × Volume (in liters)
Moles KOH = 0.4 M × 0.025 L
Moles KOH = 0.01 moles
Step 2: Calculate the amount of H2SO4 needed to neutralize the excess KOH.
Given:
- Volume of 0.28 M H2SO4 solution = 8.5 mL (or 0.0085 L)
- Molarity of H2SO4 solution = 0.28 M
To calculate the number of moles of H2SO4 used, use the formula:
Moles H2SO4 = Molarity × Volume (in liters)
Moles H2SO4 = 0.28 M × 0.0085 L
Moles H2SO4 = 0.00238 moles
Step 3: Calculate the number of moles of KOH in excess after saponification.
Excess Moles KOH = Moles KOH used - Moles H2SO4 used
Excess Moles KOH = 0.01 moles - 0.00238 moles
Excess Moles KOH = 0.00762 moles
Step 4: Calculate the saponification number.
The saponification number is defined as the number of milligrams of KOH that saponify 1 gram of fat or oil. To calculate it, use the formula:
Saponification Number = (Excess Moles KOH × Molecular Weight of KOH) / Weight of Peanut Oil (in grams)
Given:
- Molecular Weight of KOH = 56.11 g/mol
- Weight of Peanut Oil = 2 g
Saponification Number = (0.00762 moles × 56.11 g/mol) / 2 g
Saponification Number = 0.4261 moles × 56.11 g/mol / 2 g
Saponification Number ≈ 12.024 g
Since the saponification number is usually given in milligrams, convert it to milligrams:
Saponification Number ≈ 12.024 g × 1000 mg/g
Saponification Number ≈ 12024 mg
Therefore, the saponification number of peanut oil is approximately 12024 mg or 12.024 g.