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[" The pH of a solution containg "0.4MHC...

[" The pH of a solution containg "0.4MHCO_(3)" and "0.2MCO_(1)^(2)" is."],[[K_(4)(H_(2)CO_(3))=4times10^(-1)K_(10)(HCO_(3))=4times10^(-11)]],[(A)104]

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The pH of a solution containing 0.4 M HCO_(3)^(-) and 0.2 M CO_(3)^(2-) is : [K_(a1)(H_(2)CO_(3))=4xx10^(-7) , K_(a2)(HCO_(3)^(-))=4xx10^(-11)]

The pH of a solution containing 0.4 M HCO_(3)^(-) and 0.2 M CO_(3)^(2-) is : [K_(a1)(H_(2)CO_(3))=4xx10^(-7) , K_(a2)(HCO_(3)^(-))=4xx10^(-11)]

The pH of the a solution containing 0.4 M HCO_(3)^(-) is : [K_(a_(1)) (H_(2)CO_(3)) = 4 xx 10^(-7), K_(a_(2)) (HCO_(3)^(-)) = 4 xx 10^(-11)]

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) pH of solution is

Calculate [H^(o+)] in a soluton that is 0.1M HCOOH and 0.1 M HOCN. K_(a)(HCOOH) = 1.8 xx 10^(-4), K_(a) (HoCN) = 3.3 xx 10^(-4) .

Calculate [H^(o+)] in a soluton that is 0.1M HCOOH and 0.1 M HOCN. K_(a)(HCOOH) = 1.8 xx 10^(-4), K_(a) (HoCN) = 3.3 xx 10^(-4) .

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) Concentration of H_(2)PO_(4)^(-) is

The PH of 10^(-2)MNH_(4)CN solution would be- (Given that K_(a) of HCN=5times10^(-10) and K_(b) of ( ) agNH_(3) ) =2times10^(-5) )