Why is it possible for a chemical reaction to occur spontaneously even though 3.
ID: 576082 • Letter: W
Question
Why is it possible for a chemical reaction to occur spontaneously even though 3. for the reaction is positive? What is the pH of a 0.01M ammonia solution? Ka for ammonium ion is 5.62 x 10-10? 4. HUS 5. The pH of a 0.02M solution of an acid was measured at 4.6 What is the [H'] in this solution? Calculate the acid dissociation constant Ks and pKa for this acid. a. b. If the internal pH ofa muscle cell is 6.8, what is the [HPO/[H2P04] ratio in this cell? 6. 7. What is the pH of a solution containing 0.2g/L Na2CO3 and 0.2g/L NaHCOs? pKa= 10.25 for carbonic acid.Explanation / Answer
3. free energy change should be negative for a reaction to be spontaneous. From second law of thermodynamics,
G = H - TS
Spontaneity of reaction Vs enthalpy Vs entropy
When H is negative and S is positive, G will be negative over all temperatures.
When H is negative and S is negative, G will be negative at low temperatures.
When H is positive and S is negative, G will not be negative at any temperature. There would be no spontaneous reaction.
4. Self ionisation constant of water
Kw=Ka * Kb
Kw= 10-14
Kb=Kw/Ka
Here, Ka value of ammonium ion is given 5.62 * 10-10
So, Kb= 10-14/5.62 * 10-10
Kb=1.78 * 10-5
To know the pH of the solution, pOH should be known.
NH3 + H2O NH4++ OH-
0.01 0 0 0
-x (+x) (+x)
0.01-x x x
Base dissociation constant, Kb= [NH4+] * [OH-]/NH3
x * x/0.01-x
Since, Kb is too small, 0.01-x 0.01
Kb= x2/0.01= 1.7810-5
x= (0.01 * 1.7810-5) = 0.00042= [OH-]
pOH of the solution= -log [OH-]
pOH= log(0.00042)= 3.37
pH of the solution = 14pOH = 14- 3.37 = 10.63
5. pH=-log [H+]
pH= 10-4.6= 2.5 * 10-5
pKa of an acid = x2/(0.02-x)= (2.5 * 10-5)2/(0.02-2.5 * 10-5) = 3.28 * 10-8
6. pH= pKa + log ([HPO42-]/HPO4-]
pH= 6.8
log ([HPO42-]/HPO4-] = pH - pKa
[HPO42-]/HPO4-]= 10pH-10pKa
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