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CHEM 136

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A sample of 25.00 mL of 0.100 M HNO2(in a flask) is titrated with 0.150 M of NaOH solution at 25 degrees.
1) calculate the volume(Ve) of the NaOH solution needed to completely neutralize the acid in the flask.
2)calculate the pH for (a)the initial acid solution in the flask (b)the point at which 8.00 mL of the base has been added, (C) the equivalence point, and(d) the point at which 20.00 mL of the base has been added.

  • CHEM 136 - ,

    I can't do the entire titration for you.
    1. mL acid x M acid = mL base x M base.
    2. The secret to a,b,c,d, is in knowing where you are on the titration curve.
    a. Pure HNO2. Before any NzOH has been added.
    HNO2 ==> H^+ + NO2^-
    Ka = (H^+)(NO2^-)/(HNO2)
    Substitute into Ka above the following:
    H^+ = x
    NO2^- = x
    HNO2 = 0.1 - x
    Solve for x and convert to pH.

    b. Calculate moles HNO2 initially.
    Calculate moles NaOH at 8.00 mL
    Subtract moles NaOH added from moles HNO2 initially for moles HNO2 remaining unreacted. Divide by new volume to obtain M. Determine how much of the salt was formed (in moles) and divide by the new volume to find M. Then substitute into the Ka expression and solve for H^+, then convert to pH.
    c. At the equivalence point, the pH is determined by the hydrolysis of the salt.
    NO2^- + HOH ==> HNO2 + OH^-

    Kb for the NO2^- = (Kw/Ka) = (HNO2)(OH^-)/(NO2^-).
    Prepare an ICE chart, substitute into the expression I've written and solve for OH^-, convert to pOH, then to pH.
    d. The excess NaOH determines the pOH and pH. Determine how much NaOH has been added after the equivalence point, convert to moles, determine pOH, and pH.
    Post your work if you get stuck.

  • CHEM 136 - ,

    hey did you figure this out....if so what answers did you get

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