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OpenStudy (anonymous):

Q: The hydroxide ion concentration in a .157M solution of NaC3H5O2 is found to be 1.1x10^-5M Calculate the base ionization constant for propanoate ion. my attempt: [OH] = 1.1x10^-5M [NaC3H5O2] = 0.157M NaC3H5O2 --> Na = C3H5O2 1:1:1 ]C3H5O2] = [NaC3H5O2] pOH = -log[OH] pOH= 4.96 pOH + pH = 14 pH = 9.04 pH=-log[H3O] [H3O] = 10^-9.04 [H3O]= 9.12x10^-10M ka = [H3O]/[C3H5O2] I don't know what I'm doing... How do I approach this question?

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