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Determine the pH of 263 ml of solution which has [NH 4I] = 0.300 M. Kb = 1.74 × 10-5 for NH3(aq)

  • Chemistry -

    NH4+ + H2O ==> H3O+ + NH3
    Ka = (Kw/Kb) = (H3O+)(NH3)/(NH4+)
    Set up an ICE chart and substitute into the Ka expression. Solve for (H3O+) and convert to pH. You don't need the 263 mL.

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