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
Damn
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 :)
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
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.
To calculate the pH after adding different volumes of HNO3 to the solution of H2NNH2, you need to consider the stoichiometry of the reaction and the equilibrium conditions.
a) After adding 0 mL of HNO3, there is no reaction and the concentration of H2NNH2 remains at 0.100 M. To calculate the pH, you would use the expression for Kb:
Kb = [N2H5+][OH-] / [N2H4]
Since there is no reaction, the concentration of N2H5+ and OH- is equal to zero. Therefore, Kb = [OH-]^2 / [N2H4] = 3.0 x 10^-6. Solving for [OH-], you get [OH-] = sqrt(Kb * [N2H4]).
In this case, [N2H4] = 0.100 M. Plugging in the values, you get [OH-] = sqrt((3.0 x 10^-6) * 0.100) = 5.48 x 10^-4 M. To find the pH, you can use the relationship between pH and pOH: pH = 14 - pOH = 14 - log([OH-]) = 14 - log(5.48 x 10^-4) = 10.48.
b) After adding 25.0 mL of HNO3, you have a reaction between H2NNH2 and HNO3 according to the equation: H2NNH2 + HNO3 -> N2H5+ + NO3-. The stoichiometry of the reaction tells us that for every 1 mol of HNO3, 1 mol of H2NNH2 reacts to form 1 mol of N2H5+.
In this case, you have 0.100 M H2NNH2 and 0.200 M HNO3. Considering the volume of the solution and the stoichiometry, you can calculate the moles of each species:
moles of H2NNH2 = 0.100 M * 0.100 L = 0.010 mol
moles of HNO3 = 0.200 M * 0.025 L = 0.005 mol
Since the moles of H2NNH2 and HNO3 are equal (1:1 stoichiometry), they react completely.
The volume of the resulting solution is 0.100 L + 0.025 L = 0.125 L.
To find the pH, you can calculate the concentration of N2H5+ by dividing the moles by the volume of the solution:
[N2H5+] = moles of N2H5+ / volume of solution = 0.005 mol / 0.125 L = 0.040 M
To calculate the pH, you can use the relationship between pH and [H+]:
pH = -log[H+] = -log(0.040) = 8.40.
c) After adding 50.0 mL of HNO3, the stoichiometry and calculations are similar to part b). You have 0.010 mol of H2NNH2 reacting with 0.010 mol of HNO3.
The resulting volume of the solution is 0.100 L + 0.050 L = 0.150 L.
The moles of N2H5+ formed is 0.010 mol. Dividing by the volume of the solution, you get:
[N2H5+] = 0.010 mol / 0.150 L = 0.067 M
To calculate the pH, use the relationship pH = -log[H+]:
[H+] = [N2H5+]
pH = -log(0.067) = 4.82.
d) After adding 60.0 mL of HNO3, you have 0.010 mol of H2NNH2 reacting with 0.012 mol of HNO3.
The resulting volume of the solution is 0.100 L + 0.060 L = 0.160 L.
The moles of N2H5+ formed is 0.010 mol. Dividing by the volume of the solution, you get:
[N2H5+] = 0.010 mol / 0.160 L = 0.063 M
To calculate the pH, use the relationship pH = -log[H+]:
[H+] = [N2H5+]
pH = -log(0.063) = 4.80.
I hope this helps! Let me know if you have any further questions.