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To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus (H3PO3) acid the volume of 0.1 M aqueous KOH solution required is:
- a)60 mL
- b)20 mL
- c)40 mL
- d)10 mL
Correct answer is option 'C'. Can you explain this answer?
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To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus...
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To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus...
To neutralize the given 20 mL of 0.1 M aqueous solution of phosphorus (H3PO3) acid, we need to determine the volume of 0.1 M aqueous KOH solution required.
First, let's understand the reaction between phosphorus acid (H3PO3) and potassium hydroxide (KOH):
H3PO3 + KOH → K3PO3 + H2O
From the balanced chemical equation, we can see that one mole of H3PO3 reacts with three moles of KOH to produce one mole of K3PO3 and three moles of water (H2O).
Now, let's calculate the number of moles of H3PO3 in the given solution:
Moles of H3PO3 = Molarity × Volume
= 0.1 mol/L × 0.020 L
= 0.002 mol
Since the stoichiometric ratio between H3PO3 and KOH is 1:3, we need three times the amount of KOH to neutralize the acid completely.
Moles of KOH required = 3 × Moles of H3PO3
= 3 × 0.002 mol
= 0.006 mol
Now, let's calculate the volume of 0.1 M KOH solution required:
Volume of KOH solution = Moles of KOH required / Molarity of KOH
= 0.006 mol / 0.1 mol/L
= 0.06 L
Finally, we convert the volume to milliliters:
Volume of KOH solution = 0.06 L × 1000 mL/L
= 60 mL
Hence, the volume of 0.1 M aqueous KOH solution required to neutralize completely 20 mL of 0.1 M aqueous phosphorus acid is 60 mL. Therefore, the correct answer is option 'C'.
First, let's understand the reaction between phosphorus acid (H3PO3) and potassium hydroxide (KOH):
H3PO3 + KOH → K3PO3 + H2O
From the balanced chemical equation, we can see that one mole of H3PO3 reacts with three moles of KOH to produce one mole of K3PO3 and three moles of water (H2O).
Now, let's calculate the number of moles of H3PO3 in the given solution:
Moles of H3PO3 = Molarity × Volume
= 0.1 mol/L × 0.020 L
= 0.002 mol
Since the stoichiometric ratio between H3PO3 and KOH is 1:3, we need three times the amount of KOH to neutralize the acid completely.
Moles of KOH required = 3 × Moles of H3PO3
= 3 × 0.002 mol
= 0.006 mol
Now, let's calculate the volume of 0.1 M KOH solution required:
Volume of KOH solution = Moles of KOH required / Molarity of KOH
= 0.006 mol / 0.1 mol/L
= 0.06 L
Finally, we convert the volume to milliliters:
Volume of KOH solution = 0.06 L × 1000 mL/L
= 60 mL
Hence, the volume of 0.1 M aqueous KOH solution required to neutralize completely 20 mL of 0.1 M aqueous phosphorus acid is 60 mL. Therefore, the correct answer is option 'C'.
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40 ml
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To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus (H3PO3) acid the volume of 0.1 M aqueous KOH solution required is:a)60 mLb)20 mLc)40 mLd)10 mLCorrect answer is option 'C'. Can you explain this answer?
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