A solution is prepared by diluting 65.0 mL of 0.175 M methylamine, CH3NH2, to a total volume of 225 mL. Calculate the pH of the diluted solution.
Kb (CH3NH2) = 4.40 x 10-4

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  1. CH3NH2 + HOH --> CH3NH3^+ + OH^-

    Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
    Make an ICE chart and substitute into the Kb expression I've written.
    (CH3NH2) = 0.175 x (65/225) = ?
    (CH3NH3^+) = x
    (OH^-) = x
    Solve for x, convert OH^- to pH.

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