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JinKazama1:
Is 40mL answer?
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Final Answer : 40mL
Steps:
1) We know that, Basicity of H3PO3 is 2 .
Since, looking at the picture of it we easily get that.
So,
Normality of H3PO3 = 0.1*2 = 0.2N.
Volume of H3PO3 = 20mL
2) And, No. of ionizable OH- in KOH, is 1 so
Normality will be same as Molarity =0.1 *1 =0.1N
3) By Law of equivalents,
N1V1 = N2V2
=> 0.2 * 20 = 0.1 * V(req)
=> V(req) = 40mL
So, Volume of KOH required to reduce H3PO3 will be 40mL.
For structure see pic
Steps:
1) We know that, Basicity of H3PO3 is 2 .
Since, looking at the picture of it we easily get that.
So,
Normality of H3PO3 = 0.1*2 = 0.2N.
Volume of H3PO3 = 20mL
2) And, No. of ionizable OH- in KOH, is 1 so
Normality will be same as Molarity =0.1 *1 =0.1N
3) By Law of equivalents,
N1V1 = N2V2
=> 0.2 * 20 = 0.1 * V(req)
=> V(req) = 40mL
So, Volume of KOH required to reduce H3PO3 will be 40mL.
For structure see pic
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