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March 30, 2017

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Ammonia, NH3 is freely soluble in water, and has the value Kb=1.8x10^-5. If the pH of an ammonia solution is 11.4272, then what original concentration of NH3 (molarity) was used to make up the solution?

  • Chemistry - ,

    NH3 + H2O ==> NH4^+ + OH^-

    Kb = (NH4^+)(OH^-)/(NH3)
    (x)(x)/(NH3)-x = 1.8 x 10^-5
    pH = 11.4272 (where did all these places come from---we only know Kb to 2 places.)
    pH = 11.43 pOH = 14-11.43 = 2.57
    2.57 = -log(OH^-) and (OH^-) = 2.69 x 10^-3
    Substitute into Kb expression and solve for (NH3). You should get something close to 0.4 M but you need to confirm all of the above.

  • Chemistry - ,

    Thank you! I tried all the above and got 0.40 M

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