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A 30 ml sample of of 0.150 m hydrazoic acid (ka = 4.50x10^-4 ) is titrated with a 0.100 m NaOH what is the ph after the ff additions of NaOH

30ml

45ml

60ml

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  1. 1. Do you mean 0.150 m or 0.150M? You need to get your cap keys in order. I think you MUST mean M and I will work it that way. (By the way, M stands for molarity; m stands for molality.)
    a. Write the equation and balance it.
    b. Calculate mL NaOH to reach the equivalence point. You do that so you will know where each of these additions is on the titration curve; i.e., beginning, before or after the equivalence point or at the eq. pt. I calculate 30*0.150/0.1 = 45.0 mL NaOH but you should confirm that.

    c. 30 mL is between the beginning and the eq pt; therefore you have a buffered solution. Use the Henderson-Hasselbalch equation.
    d. 45 mL is at the eq pt and the pH will be determined by the hydrolysis of the salt. The cncn of the salt is 0.06 but you should confirm that. You have 30 x 0.15 = 4.5 millimols salt in 45+30 mL volume and 4.5/75 = 0.06M.
    ........N3^- + HOH ==> HN3 + OH^-
    I.....0.06..............0.....0
    C.......-x..............x.....x
    E.....0.l06-x...........x.....x

    Kb for N3^- = (Kw/Ka for HN3) = (x)(x)/(0.06-x) and solve for x = (OH^-). Convert that to pH.

    d. At 60 mL you have passed the equivalence point. Therefore, the pH is determined by the excess NaOH.
    mmols HN3 initially = 30 x 0.150 = ?
    mmols NaOH = 60 x 0.1 = ?
    Difference is excess OH^-.
    M OH^- = mmols/mL and convert that to pOH then to pH.

    I'll be happy to check your calculations if you care to post them.

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