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Standard XI
Chemistry
Solubility Equilibria
Question

$$0.00050$$ mole of $$NaH{CO}_{3}$$ is added to a large volume of a solution buffered at $$pH=8.00$$. How much material will exist in each of the three forms, $${ H }_{ 2 }{ CO }_{ 3 },H{ CO }_{ 3 }^{ - },{ CO }_{ 3 }^{ 2- }$$. $${K}_{1}$$ and $${K}_{2}$$ for $${H}_{2}{CO}_{3}$$ are $$4.5\times { 10 }^{ -7 }$$ and $$4.5\times { 10 }^{ -11 }$$ respectively.

A
$${n}_{{ H }_{ 2 }{ CO }_{ 3 }}=1.069\times {10}^{-5}$$; $${n}_{H{ CO }_{ 3 }^{ - }}=4.86\times {10}^{-4}$$; $${n}_{{ CO }_{ 3 }^{ 2- }}=2.28\times {10}^{-5}$$
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Solution
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Correct option is A. $${n}_{{ H }_{ 2 }{ CO }_{ 3 }}=1.069\times {10}^{-5}$$; $${n}_{H{ CO }_{ 3 }^{ - }}=4.86\times {10}^{-4}$$; $${n}_{{ CO }_{ 3 }^{ 2- }}=2.28\times {10}^{-5}$$

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Q1
$$0.00050$$ mole of $$NaH{CO}_{3}$$ is added to a large volume of a solution buffered at $$pH=8.00$$. How much material will exist in each of the three forms, $${ H }_{ 2 }{ CO }_{ 3 },H{ CO }_{ 3 }^{ - },{ CO }_{ 3 }^{ 2- }$$. $${K}_{1}$$ and $${K}_{2}$$ for $${H}_{2}{CO}_{3}$$ are $$4.5\times { 10 }^{ -7 }$$ and $$4.5\times { 10 }^{ -11 }$$ respectively.
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Q2
If 0.00050 mol NaHCO3 is added to 1 litre of a buffered solution of pH 8, then how much material will exist in each of the three forms H2CO3,HCO−3 and CO2−3?
For H2CO3, K1=5×10−7; K2=5×10−13

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Q3
H2CO3 ionises as
H2CO3⇌H++HCO−3;K1=4.3×10−7
HCO−3⇌H++CO2−3;K2=5.6×10−11
Calculate the degree of hydrolysis and pH value of 0.12M Na2CO3 solution.
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Q4
Carbonic acid, H2CO3, is a diprotic acid for which K1=4.2×10−7 and K2=4.7×10−11. Which solution will produce a pH closest to 9?
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Q5
pH of 0.1 M NaHCO3 if K1=4.5×10−7,K2=4.5×10−11.
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