1. Questions

chemistry

hi

kinda stuck :)

100.0 ml of 0.100 M H2NNH2 (kb =3.0 x10^-6) is titrated by 0.200M

calculate PH after adding the following volumes of HNO3

a 0
b 25.0 ml
c 50.0 ml
d 60.0 ml

-----------------------------------------

A)

N2H4 + HNO3 --> N2H5+ + NO3-

.100 0 0


3.0 x 10-6= x2/.100

so i got PH 10.74 answer is 10.48

B)

N2H4 + HNO3--> N2H5+ + NO3-

.01 mol .005mol

-.005 .005
-----------------------------------
.005 Mol .005 mol
--------- -------
.125 L .125 L

as you can see they cancel so i did the ph stuff

PH=8.48

c)N2H4 + HNO3 --> N2H5+ + NO3-
.01 mol .01

-.01 -.01 .01
---------------------------
0 0 .01 MOL

n2h5 + h20----n2h4 + h30

.08 M

1.0 x10-14/3.0 x10-6=3.3 x10-9

ka= 3.3 x10-9 * x2/.08

i got 4.79 answer is 4.83

D)N2H4 + HNO3 --> N2H5+ + NO3-
.010mol .012 mol
-.010 .010Mol
-----------------------------------
0 .002 mol .010
----- ---
.160 .160

--------------------------------
.13M .063M

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Jim_R

  1. I think i figure them out

    But i still don't understand them conceptually.

    like in D

    We have an excess of Hno3 and H2nnh2 is the limiting reagent. Yet we use this formula

    Hno3 + h20---etc.

    What happend to the H2NNH3

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    Jim_R

  2. I agree with your answer to (a) of 10.74.
    I agree with b and c.
    For D,
    H2NNH2 + HNO3 ==> H2NNH3+^ + NO3^-

    You had 0.1 L x 0.1 M = 0.01 moles H2NNH2.
    You added 0.06 L x 0.2 M = 0.012 moles HNO3.
    You had an excess of 0.002 mols HNO3 (just has you have it).
    So there is an excess of HNO3 as well as 0.01 moles of the salt (H2NNH3^+) present. Which determines the pH, the hydrolysis of the salt (which will occur to a slight extent) or hydrogen ion formation from a STRONG acid like HNO3. Of course, the HNO3 will do it so the final pH is
    0.002 moles/0.0160 L = 0.0125 M (H3O^+) and that gives a pH of 1.90. The salt is still there and it is still hydrolyzing (just as it did for the 25 mL and 50 mL additions) BUT hydrolysis accounts for so much less than an excess of a strong acid such as HNO3.

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    DrBob222



  4. thanks dr. bob

    I should be paying you 400 dollar rather than my college

    Can you help me with my labs. Im having some problems since i haven't done em before. I forgot how to calculate titrations last one so i kinda messed up :)

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    Jim_R

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