help please ASAP!!! 1) A 0.040 M solution of a weak acid (HA) has a pH of 4.70.
ID: 587800 • Letter: H
Question
help please ASAP!!!
1)
A 0.040 M solution of a weak acid (HA) has a pH of 4.70. What is the pKa of the acid?
6.00
8.00
2.89
4.12
5.34
2)
The following reaction has been studied at 25oC.
2BrCl Br2 + Cl2 Kc = 0.141 at 25oC
If the initial concentration of chlorine is 0.0300 M and of bromine monochloride is 0.0200 M, what is the equilibrium concentration of bromine?
3)
Carbon monoxide and chlorine combine in an equilibrium reaction to produce the highly toxic phosgene (COCl2):
CO(g) + Cl2(g) COCl2(g) Kc = 248
What will happen when the reactants and product are combined with the following concentrations: [CO] = [Cl2] = 0.0200 M; [COCl2] = 0.0992 M
- The reaction will proceed to the left.
- The reaction is at equilibrium, and no change in concentrations will occur.
6.00
8.00
2.89
4.12
5.34
2)
The following reaction has been studied at 25oC.
2BrCl Br2 + Cl2 Kc = 0.141 at 25oC
If the initial concentration of chlorine is 0.0300 M and of bromine monochloride is 0.0200 M, what is the equilibrium concentration of bromine?
3)
Carbon monoxide and chlorine combine in an equilibrium reaction to produce the highly toxic phosgene (COCl2):
CO(g) + Cl2(g) COCl2(g) Kc = 248
What will happen when the reactants and product are combined with the following concentrations: [CO] = [Cl2] = 0.0200 M; [COCl2] = 0.0992 M
- The reaction will proceed to the left.
- The reaction is at equilibrium, and no change in concentrations will occur.
Explanation / Answer
1)
we have below equation to be used:
pH = -log [H+]
4.7 = -log [H+]
log [H+] = -4.7
[H+] = 10^(-4.7)
[H+] = 1.995*10^-5 M
Lets write the dissociation equation of HA
HA -----> H+ + A-
4*10^-2 0 0
4*10^-2-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1.995*10^-5*1.995*10^-5/(0.04-1.995*10^-5)
Ka = 9.958*10^-9
we have below equation to be used:
pKa = -log Ka
= -log (9.958*10^-9)
= 8.00
Answer: 8.00
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