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Joined: Mar 23 2006
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Oct 22 2014 09:05am
Hey guys I have some chemistry problems I could use some step by step help to the answers with preferably by tomorrow morning. I don't have much fg but I'd be happy to divvy it out for some help. I've worked out a few of them but not 100% they're right.

1. For the reaction:
C(s) + 2H2(g) <-> CH4(g) at T=1000C
The following conditions apply: 0.100 mol CH4 and an excess of C(s) were brought together in a 4.15L reaction vessel. At equilibrium, the total pressure in the vessel was 3.65 atm.
i. Write the equilibrium expression for the reaction:
ii. Calculate the equilibrium constant, Kc:
iii. What is the value of Kp?
iv. What are the equilibrium concentrations of each reactant / product?

2.
i. How many moles of NaC2H3O2 must be added to 2.0 L of 0.10M HC2H3O2 to give a buffer solution pH=5? (Ignore any volume change due to the addition of solid NAC2H3O2), the Ka = 1.7 x 10^-5 for HC2H3O2.
ii. If 100 mL of 0.10M NaOH was added to the above solution, what would be the resultant pH of the solution?

3. Methylammonium chloride, CH3NH3Cl, is a salt of methyl amine, CH3NH2. A 0.10M solution of this salt has a pH of 5.82.
i. Write the dissociation reaction for the salt CH3NH3Cl and calculate the ka for the reaction:
ii. What would the kc value be for methyl amine (CH3NH2) dissolved in water?

4. Calculate the pH of 0.26M solution of NH4NO3 (Kb for NH3 = 1.8 x 10^-5)

5. Two 0.7182g samples of an unknown monoprotic weak acid are individually dissolved in 25 mL of deionized water. One of the samples was titrated with NaOH and required 40.83 mL of 0.1012M NaOH to reach the color endpoint. Once the endpoint was achieved, the second 25 mL portion of the weak acid was added to the first sample and mixed thoroughly. The pH of the mixed solution is found to be 3.84. Determine:
a) molar mass of weak acid
b) ka for weak acid

This post was edited by Hysteria on Oct 22 2014 09:26am
Member
Posts: 4,808
Joined: Mar 23 2006
Gold: 36.49
Oct 22 2014 03:56pm
I think I've got some of these if anyone can confirm.

1.
a) [CH4] / [H2]^2
b ) Kc = [0.100] / [0.100]^2 = 1.0x10^1
c) Kp = Kc(RT)^change in n
Kp = 10(0.0821)(1273)^1 = 1.0 x 10^4
d) Use ICE table for this? Not sure if I have this one set up right.


2.
a) pKa = -logKa
-log(1.7x10^-5) = 4.76
pH=pka + log(A-/HA)
5 = 4.76 + +log (A-/.10)
10^.24 = 10^log(A-/.10)
1.74 = (A-/.10)
.174 = A-
2L x .174 = .387 moles
Thanks to SixFour for this one.

b ) Going from there I did...
pH = 4.77 + log(.347/.10)
pH = 4.77 + log(3.47)
pH = 4.77 + .54
pH = 5.31?

4. Calculate the pH of 0.26M solution of NH4NO3 (kb for NH3 = 1.8 x 10^-5)
a) first find Ka
Ka x Kb = Kw
Ka x 1.8x10^-5 = 1 x 10^-14
Ka = 5.5x10^-10

b ) ICE table
NH4NO3 -> NH4 + NO3
I .26 0 0
C -x +x +x
E .26-x x x

c)
x^2/.26 - x = 5.5x10^-10
(assume x is so small/negligible)
x = 1.2x10^-5
pH = -log(1.2x10^-5)
pH = 4.92?

This post was edited by Hysteria on Oct 22 2014 04:00pm
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