The equilibrium expression for Kp for the reaction below is _____. A) 3PO_2/2PO_

Question

The equilibrium expression for Kp for the reaction below is _____. A) 3PO_2/2PO_3 B) 2PO_3/3PO_2 C) 3PO_3/2PO_2 D) PO^2_3/PO^3_2 E) PO^3_2/PO^2_3 What is the conjugate base of OH^-? A) O_2 B) O^- C) H_2O D) O^2- E) H_3O^+ In which aqueous system is PbI_2 least soluble? A) H_2O B) 0.5 M HI C) 0.2 M HI D) 1.0 M HNO_3 E) 0.8 M KI The equilibrium constant for the gas phase reaction is N_2(g) + 3H_2(g) 2NH_3(g) is K_eq = 4.34 times 10^-3 at 300 degree C. At equilibrium, _____. A) products predominate B) reactants predominate C) roughly equal amounts of products and reactants are present D) obly products are present E) only reactants are present Which of the following aqueous solutions has the lowest [OH^-]? A) a solution with a pH of 3.0 B) a 1 times 10^-4 M M solution of HNO_3 C) a solution with a pOH of 12.0 pure water E) a 1 times 10^-3 M solution of NH_4Cl The equilibrium-constant expression for the reaction is given by Ti(s) + 2Cl_2 (g) TiCl (l) is given by A) [TiCl (l)]/[Ti(s)][Cl_2 (g)]^2 A 0.1 M solution of _____ has a pH of 7.0. A) Na_2S B) KF c) NaNO_3 D) NH_4Cl E) NaF What is the concentration (in M) of hydroxide ions in a solution at 25.0 degree C with pH = 4.282? A) 4.28 B) 9.72 C) 1.91 times 10^-19 D) 5.22 times 10^-5 E) 1.66 times 10^4

Explanation / Answer

1.for the reaction , 2O3<---> 3O2, Kp = [PO2]3/ [PO3]2 ( d is correct)

2.   when an acid   loses H+, it becomes conjuagte base of acid.

for example, HX losses H and we will have X-, conjugate base.

So OH- loses H and its conjuate base is O-2

3. PbI2--------> Pb+2 + 2I-

As per Lechatlier principle, the presence of either Pb+2 and I- reduces the solubiliy of PbI2. Hence 0.8M KI will have greater concentration of I- ions compared to 0.5M HI or 0.2M HI, least solubiltiy is there in 0.8MKI

4. for the reaction, Kp= equilibrium constant = [NH3]2/ [N2][H2]3= 4.34*10-3, at 300 deg.c

since KP value is less, denominator is more hene equilibrium is dominated by reactants. ( B is correct)

5. pH= -log [H+], [H+]= 10-(pH)

pOH= 14-pH and [OH-] =10(-OH)

lower pH means higher pOH and lowe the OH- concentrations.

pH =3, pOH= 11

for HNO3 ( strong acid), pH= 4, pOH= 14-4=10

pOH=12 for third solution

pH= 7 for distillated water, pOH= 14-7= 7

1*10-3 M NH4Cl , NH4Cl+ H2O ------->NH3+ OH-

Ka= [NH3+] [OH-]/ [NH4Cl] [H2O], Ka for NH3= 5.6*10-10

let x= drop in concentration of NH4Cl to reach equilibrium. At equilbrium, [ NH3] =[OH-]=x and [NH4Cl]= 1*10-3-x

Ka= x2/(1*10-3-x)= 5.6*10-10, when solved using excel, [OH-]=7.48*10-7, pOH= 6.12

lowest poH is highest with a solution of

poH= 12, hence lowest OH- ( c is correct)

6. since the activity of liquid and solid is unity,

K= 1/[Cl2 (g)]2 =

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