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anonymous

  • one year ago

determine the ph of 0.015 M solution of h2so4. the dissociation occurs in two steps ka1 is extremely large ka2 is 1.2*10^-2

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  1. aaronq
    • one year ago
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    Since the first Ka is extremely large, we assume that the dissociation of the first proton goes to completion, meaning \([H^+]=[H_2SO_4]\) after we use the equilibrium expression for the second proton, \(HSO_4^-\rightleftharpoons H^++SO_4^{2-}\) \(K_{a2}=\dfrac{[H^+][SO_4^{2-}]}{[HSO_4^-]}\) we also know that \([HSO_4^-]\) is equal to initial concentration of \(H_2SO_4\)

  2. aaronq
    • one year ago
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    Add up the concentrations of \(H^+\) then use them to find the final pH

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