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To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the value of 0.1 M aqueous KOH solution required is [2004]
- a)40 mL
- b)20 mL
- c)10 mL
- d)60 mL
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorou...
Solution:
The balanced chemical equation for the reaction between H3PO3 and KOH is:
H3PO3 + 3KOH → K3PO3 + 3H2O
From the balanced chemical equation, it is clear that 3 moles of KOH are required to neutralize 1 mole of H3PO3.
Number of moles of H3PO3 in 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3) can be calculated as follows:
Number of moles of H3PO3 = Molarity × Volume (in liters)
= 0.1 × 20/1000
= 0.002 moles
Since 3 moles of KOH are required to neutralize 1 mole of H3PO3, the number of moles of KOH required to neutralize 0.002 moles of H3PO3 is:
Number of moles of KOH = 3 × 0.002
= 0.006 moles
The volume of 0.1 M aqueous KOH solution required to provide 0.006 moles of KOH can be calculated as follows:
Volume of 0.1 M aqueous KOH solution = Number of moles of KOH / Molarity of KOH
= 0.006 / 0.1
= 0.06 L
= 60 mL
Therefore, the correct option is (A) 40 mL.
The balanced chemical equation for the reaction between H3PO3 and KOH is:
H3PO3 + 3KOH → K3PO3 + 3H2O
From the balanced chemical equation, it is clear that 3 moles of KOH are required to neutralize 1 mole of H3PO3.
Number of moles of H3PO3 in 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3) can be calculated as follows:
Number of moles of H3PO3 = Molarity × Volume (in liters)
= 0.1 × 20/1000
= 0.002 moles
Since 3 moles of KOH are required to neutralize 1 mole of H3PO3, the number of moles of KOH required to neutralize 0.002 moles of H3PO3 is:
Number of moles of KOH = 3 × 0.002
= 0.006 moles
The volume of 0.1 M aqueous KOH solution required to provide 0.006 moles of KOH can be calculated as follows:
Volume of 0.1 M aqueous KOH solution = Number of moles of KOH / Molarity of KOH
= 0.006 / 0.1
= 0.06 L
= 60 mL
Therefore, the correct option is (A) 40 mL.
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Community Answer
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorou...
H3PO4 has two H+ to donate so on equalate their eq.wt.
2*20*0.1=x*1*0.1
x=40ml
2*20*0.1=x*1*0.1
x=40ml
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To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the value of 0.1 M aqueous KOH solution required is [2004]a)40 mLb)20 mLc)10 mLd)60 mLCorrect answer is option 'A'. Can you explain this answer?
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