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Calculate the pH of 0.05 M KHC(8)H(4)O(4...

Calculate the `pH` of `0.05 M KHC_(8)H_(4)O_(4)`
`H_(2)C_(8)H_(4)O_(4) + H_(2)O hArr H_(3)O^(o+) + HC_(8)H_(4)O_(4)^(Theta) pK_(a_(1)) = 2.94`
`HC_(8)H_(4)O_(4)^(Theta) + H_(2)O hArr H_(3)O^(o+) + C_(8)H_(4)O_(4)^(2-) pK_(a_(2)) = 5.44`

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The correct Answer is:
4.19
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Calculate the pH of 0.05 M KHC_(8)H_(4)O_(4) H_(2)C_(8)H_(4)O_(4) + H_(2)O hArr H_(3)O^(o+) + HC_(8)H_(4)O_(4)^(o+) pK_(a_(1)) = 2.94 HC_(8)H_(4)O_(4)^(Theta) + H_(2)O hArr H_(3)O^(o+) + C_(8)H_(4)O_(4)^(2-) pK_(a_(2)) = 5.44

In the equilibrium HClO_(4) + H_(2)O hArr H_(3)O^(+) + ClO_(4)^(-)

H_(3)PO_(4)+H_(2) Leftrightarrow H_(3)O^(+)+H_(2)PO_(4)^(-),pK_(1)=2.15 H_3PO_(4)^(-)+H_(2)O Leftrightarrow H_(3)O^(+)+HPO_(4)^(2-), pK_(2)=7.20 Hence pH of 0.01 M NaH_(2)PO_(4) is

H_(4)underline(S_(2))O_(6)+H_(2)O to H_(2)SO_(3)+H_(2)SO_(4)

H_(4)underline(P_(2))O_(6)+H_(2)O to H_(3)PO_(3)+H_(3)PO_(4)

H_(4)underline(P_(2))O_(8)+H_(2)O to H_(3)PO_(4)+H_(2)O_(2)

H_(4)underline(B_(2))O_(6)+H_(2)O to H_(3)BO_(3)+H_(2)O_(2)

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