Archive for November, 2008
Free chemistry Tutor Homework Help
Consider the titration of 20.0 mL of 0.0800 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HClO4.? Calculate the pH after the following volumes of titrant have been added:
(a) 0.0 mL
(b) 4.0 mL
(c) 8.0 mL
Answer
(a). Since no titrant has been added to the analyt so we only have 20.0 mL of 0.0800 M (CH3)3N thus the pH of this solution is calculated using this equation:
[OH-] = (Kb x M)1/2
[OH-] = (6.4.10-5x 0.0800)1/2
[OH-] = 2.26×10-2
pOH = - log 2.26×10-2 = 2.64
pH = 14- 2.64 = 11.36
b. 20.0 mL of 0.0800 M (CH3)3N is equal to 1.6 mmol of (CH3)3N and we add 4.0 mL HClO4 0.100 M that equal to 0.4 mmol. Thus 1.6 mmol (CH3)3N to be react with 0.4 mmol HClO4 this will be produce 0.4 mmol of (CH3)NHCLO4. The solution contains base and its salt, so this will be form a buffer solution
The balance reaction after with mole after reaction occur:
(CH3)3N + HClO4 -> (CH3)3NHClO4
1,2 mmol—0 mmol———0.4 mmol
We can use the henderson-Hasselbalch equation to calculate pH of the buffer solution
pH = pKa + log (base/salt)
Ka for (CH3)3N = Kw/Kb = 1×10-14/6.4.10-5 = 1.56.10-10
pKa = - log Ka = - log(1.56.10-10) = 9.8
pH = 9.8 + log (1.2 mmol/0.4 mmol) = 10
c. Using the same way that we use to solve the b problem we get:
(CH3)3N + HClO4 -> (CH3)3NHClO4
0.8 mmol—-0 mmol—-0.8 mmol
pH = 9.8 + log (0.8 mmol/0.8 mmol) = 9.8
November 5th, 2008
Free Chemistry Tutor Homework help
How many milliliters of 0.861 M HBrO4 are needed to titrate each of the following solutions to equivalence point?
- 75.0 mL of 1.46 M LiOH
- 48.1 mL of 1.55 M NaOH
- 281.0 mL of a solution that contains 44.2 g of RbOH per liter
Answer
For titration problem, there is a formula that you can use to solve titration problem. This formula is uneversal for all kind of titration, and the formula is:
V x N (acid) = V x N (base)
V is the volume and N is the normality. For the problem above we need to calculate the normality of all the solution. Just for remember, to change from molarity into normality for acid or base you just multiply the molarity with the number of proton (acid) or the number of OH- for base.
0.861 M HBrO4 = 0.861 N HBrO4
1.46 M LiOH = 1.46 N LiOH
1.55 M NaOH = 1.55 N NaOH
Normality of 281.0 mL of a solution that contains 44.2 g of RbOH are:
mole = 44.2 / 102.48 = 0.431 mol
M = 0.431 mol/0.281 L = 1.53 M = 1.53 N
1. 75.0 mL of 1.46 N LiOH
75.0 x 1.46 N = 0.861 N x V
V = 127,2 mL
2. 48.1 mL of 1.55 N NaOH
48.1 x 1.55 N = 0.861 N x V
V = 86.6 mL
3. 281.0 mL of 1.53 N RbOH
281.0 x 1.53 N = 0.861 N x V
V = 499.3 mL
November 5th, 2008
Free Chemistry Tutor Help Problem
Calculate the enthalpy change for the formation of PbO from Pb to O2 using this reaction
Pb+CO —-> PbO + C ΔH = -106.8KJ
2C + O2 —–> 2CO ΔH = -221.0KJ
Answer
This is the examples of Thermochemistry that can be solve using Hess Law. rearrange the known reaction into the reaction that the problem wants. You need to calculate this reaction:
PbO -> Pb + 1/2 O2 ΔH= ?
First step we want PbO on the left or to be as reactant from the first reaction above PbO is on the right thus we should reverse this reaction
PbO + C -> Pb + CO ΔH= 106.8 KJ
second we need to reverse the second reaction above thus we get O2 on the left and divide the reaction with two to get the reaction coefficient of O2 become 1/2
CO -> C + 1/2O2 ΔH= 110.5 KJ
the reaction become
PbO + C -> Pb + CO ΔH= 106.8 KJ
CO -> C + 1/2O2 ΔH= 110.5 KJ
——————————————–
PbO -> Pb + 1/2 O2 ΔH= 217.3 KJ
November 5th, 2008
Free Chemistry Homework Help
A 18.0 L gas cylinder has been filled with 5.00 moles of gas. You measure the pressure to be 4.30 atm.? What is the temperature inside the tank?
Answer
A particular of gas is described by its pressure, volume, temperature, and the number of moles. Knowledge of any of three of these properties is enough to completely define the state of the gas, since the fourth properties can then determine from the equation of ideal gas law.
PV = nRT
R is universal gas constant 0.08206 L.amt/K.mol
P is the pressure in atm
V is the volume in Liter
n is the moles
T is temperature in Kelvin
4.30x 18.0 L = 5 x 0.08206 x T
T = 188.6 K
November 5th, 2008