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3 years ago

The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?

Chemistry

1 answer:

svlad2 [7]3 years ago

6 0

Answer: Dissociation constant of the acid is $10^{-11}$ .

Explanation: Assuming the acid to be monoprotic, the reaction follows:

$HA\rightleftharpoons H^++A^-$

pH of the solution = 6

and we know that

$pH=-log([H^+])$

$[H^+]=antilog(-pH)$

$[H^+]=antilog(-6)=10^{-6}M$

As HA ionizes into its ions in 1 : 1 ratio, hence

$[H^+]=[A^-]=10^{-6}M$

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

$[HA]=[HA]-[H^+]$

$[HA]=0.1M-10^{-6}M$

$[HA]=0.09999M$

Dissociation Constant of acid, $K_a$ is given as:

$K_a=\frac{[A^-][H^+]}{HA}$

Putting values of $[H^+],[A^-]\text{ and }[HA]$ in the above equation, we get

$K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}$

$K_a=1.0001\times 10^{-11}M$

Rounding it of to one significant figure, we get

$K_a=1.0\times 10^{-11}M\approx 10^{-11}M$

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Answer:

The balanced reaction in the acidic medium is:

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Balance oxygen by adding water :

$CN^-+H_2O\rightarrow CNO^-$

Now , balance hydrogen by adding hydrogen ions on the side where hydrogen are less.

$CN^-+H_2O\rightarrow CNO^-+2H^+$

Add electrons to balance charge where positive charge is more.

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Balance oxygen by adding water :

$MnO4^-\rightarrow MnO_2+2H_2O$

Now , balance hydrogen by adding hydrogen ions by adding them on the side where hydrogen are less.

$MnO4^-+4H^+\rightarrow MnO_2+2H_2O$ ..[2]

Add electrons to balance charge where positive charge is more.

$MnO4^-+4H^++3e^-\rightarrow MnO_2+2H_2O$ ..[2]

3 × [1] + 2 × [2]

The balanced reaction in the acidic medium is:

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givi [52]

Answer: The order from the Most energy released to most Energy Absorbed Is given as 2---> 4--->,3-->---> 1

B)-61.9 kJ

Explanation:

The change in the internal energy of a system is positive if the reaction absorbs energy and negative if the reaction releases energy. For a system to cause an increase in volume, it must have very high energy built up to be released.

1. Surroundings get colder and the system decreases in volume. Here, the surrounding absorbs energy resulting in positive ΔE

2. Surroundings get hotter and the system expands in volume. Here energy is released causing the system to be negative

3. Surroundings get hotter and the system decreases in volume. Although there is a decreased volume, the system is negative because it releases energy

4. Surroundings get hotter and the system does not change in volume. System is negative because it releases energy even thgoygh there is no change in volume

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b) Using

ΔE = q+ w from 1st law of thermodynamics

ΔE= ΔH - P ΔV

gIven

ΔH = -75.0KJ

volume= A change from 5.0L TO 2.0L = Final volume - initial volume = 2-5= -3.00L

P= 43.0atm

ΔE= ΔH - P ΔV

P ΔV = 43 atm x -3 = -129L.atm

We first convert L-atm to Joules.

1 L-atm = 101.325 Joules.

129L.atm = 129 x 101.325 = - 13071 J

to KJ becomes

13071/1000 = - 13.071KJ

Recall ΔE= ΔH - P ΔV and putting values

ΔE = -75.0 - (-13.071 KJ)= -75.0 kJ + 13.071 kJ = -61.9 kJ

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