Can someone please help me step by step with this question? Given a diprotic acid, H2A,...

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Image for Given a diprotic acid, H2A, with two ionization constants of K = 2.74x 10^-4 and Ka2 = 5.24x 10-11, calculate

Can someone please help me step by step with this question?

Answer 1

Answer #1

0.143 M SOLUTION H2A:it is a weak acid dissociates H^+FROM H2A>>(H^+)FROM WATER

Sqrt(Ka1x(H2A))

Sqrt((2.74x10^-4)(0.143))

Sqrt=((2.74x1/10000)(o.143))

sqrt0.000039182

0.00625.

PH=-LOG(H30^+)

-LOG (O.00625)=

H2A=

O.143-X=

0.143-0.00625=0.13675

A2^-=

Ka2=5.24x10^-11

b) o.143 m NaHA

(H^+)=Sqrt(K2(HA^-)+K.W)/(H(HA^-)/K1) AND CONVERT TO PH

Answer 2

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