The exothermic reaction: H_2(g) + I_2(g) 2 HI(g) It allowed reach equilibrium at

Question

The exothermic reaction: H_2(g) + I_2(g) 2 HI(g) It allowed reach equilibrium at a certain temperature in a cylinder equipped with a movable piston. The volume of the system in then increased by pulling the piston outward. As a result, the equilibrium: The exothermic reaction: H_2(g) + I_2(g) 2 HI(g) is allowed to reach equilibrium at a certain temperature in a cylinder equipped with a movable piston. The temperature of the reaction vessel is then lowerd. As a result, the equilibrium: The acid dissociation constant, K_a, of a certain acid, HA, is 2.95 10^-5. The pH of a solution of 0.864 M of this acid in water is The acid dissociation constant. K_a, of a certain acid is 1.23 10^-5. The pH of a solution of 0.850 M of NaA in water is: The acid dissociation, K_a, of a certain acid, HA, is 4.68 10^-6. The pH of a combined solution of 0.0325 M HA and 0.00963 M NaA is In an acidic solution, the following redox reaction takes place: MnO_4^-(aq) + H_2C_2O_4(aq) rightarrow CO_2(g) + Mn^2+(aq) The corresponding balanced equation is:

Explanation / Answer

19)

the reaction can be written as

H2 + I2 --> 2HI + heat

according to Lechatlier principle

the equilibrium will shift in a direction to counter the change

as the volume is increased

the equilibrium will shift to decrease the volume

so

the equilibrium will shift in the direction with less moles

in this case , there are equal moles on both sides

so

no change

c) will not shift

20)

H2 + I2 --> 2HI + heat

the temperature is lowered

so

To counter the change, the system will shift to produced more heat

so

the equilibrium will shift towards right side

b) will shift to the right

21) for weak acids

[H+] = sqrt ( Ka x C)

so

[H+] = sqrt ( 2.95 x 10-5 x 0.864)

[H+] = 5.048 x 10-3

now

pH = -log [H+]

so

pH = -log 5.048 x 10-3

pH = 2.30

so

the pH is 2.30

22)

NaA is a weak base

for weak bases

[OH-] = sqrt ( Kb xC)

now

Kb = kw / Ka

so

Kb = 10-14 / 1.23 x 10-5 = 8.13 x 10-10

now

[OH-] = sqrt ( 8.13 x 10-10 x 0.85)

[OH-] = 2.63 x 10-5

pOH = -log [OH-]

pOH = -log 2.63 x 10-5

poH = 4.58

now

pH = 14 - 4.58

pH = 9.42

so

the pH is 9.42

the answer is g) 9.42

23) HA and NaA form a buffer

for a buffer

pH = pKa + log [ salt / acid ]

pH = -log Ka + log [ NaA / HA]

pH = -log 4.68 x 10-6 + log [ 0.00963 / 0.0325]

pH = 4.80

so

the pH of the solution is 4.80

24) the balanced equation is

5H2C2O4 + 2MnO4- + 6H+ --> 10CO2 + 2Mn2+ + 8H2O

25)

the balanced reaction is

2Mn04- + 4H20 + 3Cu ---> 3 Cu(OH)2 + 2 Mn02 + 2 OH-

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.