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# Buffer Solution and Entropy

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1-A 1.0-L buffer solution is 0.105M in HNO 2 and 0.145M in NaNO 2. Determine the concentrations of HNO 2 and NaNO 2 after addition of 1.4g NaOH .Express your answers using three significant figures separated by a comma.

2- A 1.0-L buffer solution is 0.105M in HNO 2 and 0.145M in NaNO 2. Determine the concentrations of HNO 2 and NaNO 2 after addition of 1.4g HI. Express your answers using three significant figures separated by a comma.

3- Which processes are nonspontaneous?
a bike going up a hill
a meteor falling to Earth
obtaining hydrogen gas from liquid water
a ball rolling down a hill

4-predict the sign of ΔS sys for each chemical reaction. ΔS sys >0 or ΔS sys <0
2KMnO 4 (s)→K 2 MnO 4 (s)+MnO 2 (s)+O 2 (g)
CH 2 =CH 2 (g)+H 2 (g)→CH 3 CH 3 (g)
Na(s)+1/2Cl 2 (g)→NaCl(s)
N 2 (g)+3H 2 (g)→2NH 3 (g)

5-Why does a gas have greater entropy than the corresponding solid? Choose from:

The entropy change of sublimation (solid going directly to gas) is positive.
There are more available types of kinetic energy in a gas. In a solid, the molecules can only vibrate.
A gas expands to fill its container, while a solid maintains a constant shape.
Particles in a solid move slower than in a gas.

##### Solution Summary

This solution clearly explains the basic concepts pertained to buffer solution and entropy, offers sound rationale and shows answers to some assorted problems in a step-wise manner.

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1-A 1.0-L buffer solution is 0.105M in HNO 2 and 0.145M in NaNO 2 .Determine the concentrations of HNO 2 and NaNO 2 after addition of 1.4g NaOH .Express your answers using three significant figures separated by a comma.
In this example HNO2 is weak acid and NaNO2 is the conjugate base.
NaOH is a strong base and dissociates completely. Therefore, adding the strong base results in 1.4 / 40 [MW of NaOH is 40] = 0.035 moles of OH- ions into the solution.
All of the OH- ions will react with the nitrous acid (HNO2) to form nitrite ions (NO2-):
HNO2 + OH- NO2- + H2O
Thus, when NaOH is added, the number of HNO2 molecules in the solution will decrease, but the number of NO2- molecules in the solution will increase. In this case, 0.035 moles of HNO2 will be used up to neutralize the 0.035 moles of OH-, and this will form 0.035 moles of NO2- in the solution.
At the end of the reaction:
Moles of HNO2 = 0.105 - 0.035 = 0.07 moles i.e. 0.07 M concentration
Moles of NO2- = 0.145 + 0.035 = 0.18 moles i.e. 0.18 M ...

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