what is the PH of a solution prepared by adding enough water to 15.00g of NaNO2 to make 275ml of solution

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asked by Chuck
  1. Calculate the molarity.
    M = mols/L and mols = grams/molar mass
    I get 15.00/69 = 0.217 mols
    and M = 0.217/0.275 L = 0.790 M. You need to confirm this. I just estimated the molar mass.
    The pH is determined by the hydrolysis (reaction with water) of the nitrite ion.

    NO2^- + HOH ==> HNO2 + OH^-

    Kb = Kw/Ka = (HNO2)(OH^-)/(NO2^-) = ??

    You know Kw and ka for HNO2 so you can calculate Kb.
    Now wet up an ICE table.

    I = Initial concentration:
    C = change in concentration:
    E = equilibrium concentration.

    (NO2^-) = 0.790 M
    (HNO2) = 0
    (OH^-) = 0

    (OH^-) = +y
    (HNO2) = +y
    (NO2^-) = -y

    (NO2^-) = 0.790 - y
    (HNO2) = 0 + y = y
    (OH^-) = 0 + y = y

    Now substitute the equilibrium values above into the Kb expression above, and solve for y. That will be equal to the (OH^-), change that to pOH, then subtract from 14 to obtain pH.
    Post your work if you get stuck.

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  2. thank you so much!

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    posted by Chuck

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