Given the following 2 reactions, what is the value of K_c for Reaction 3 and doe

Question

Given the following 2 reactions, what is the value of K_c for Reaction 3 and does this indicate a reactant or product favored reaction at equilibrium? Reaction 1: H_2S (aq) + H_2O (1) HS^- (aq) + H_3O^+ (aq) K_2 (H_2S) = 8.9 Times 10^-8 Reaction 2: HNO_2 (aq) + H_2O (1) NO_2^- (aq) + H_3O^+ (aq) K_2 (HNO_2) = 4.6 Times 10^-4 Reaction 3: HNO_2 (aq) + HS^+ (aq) H_2S (aq) + NO_2^+ (aq) K_c = ? K_c = 2.4 Times 10^10, product favored K_c = 1.9 Times10^-4, reactant favored K_c = 4.1 Times 10^-11, reactant favored K_c = 5.2 Times 10^3. product favored K_c = 2.4 Times 10^-4, reactant favored Is a solution that has a pH of 6.78 acidic, basic, or natural at a temperature of 75 degree C where K_w = 2.00 Times 1o^-13? Neutral basic acidic impossible to tell from the information given I'm just guessing What is the concentration of [H_3O^+] in a 0.245 M solution of NH_4CI? K_b(NH_3)= 1.76 Times 10^-5 K_a (HC1)= 1 Times 10^8 2.07 Times 10^-3 M 0.245 M l.18 Times 10^-5 M 4.84 TIMES 10^-12M 8.47 Times 10^-10 M Which of the following is not a strong acid? HI HBr HNO_3 HCIO_4 H_2SO_3

Explanation / Answer

11)

For given reactions,

Reaction 1: H2S (aq.) + H2O (l) <--------> HS (aq.) + H3O+ (aq.)

Ka (H2S) = [HS][H3O+] / [H2S] ……… (1) …… (since [H2O] remains almost constant).

Reaction 2: HNO2 (aq.) + H2O (l) <--------> NO2- (aq.) + H3O+ (aq.)

Ka (HNO2) = [NO2-][ H3O+]/[HNO2] ………. (2)

Now for the,

Reaction 3: HNO2 (aq.) + HS (aq.) <----------> H2S (aq.) + NO2- (aq.)

Equilibrium constant Kc can be defines as,

Kc = [H2S][NO2-]/[ HNO2][ HS] ………….. (3)

Let us rearrange this equation,

Kc = {[NO2-]/[ HNO2]} / {[ HS] /[H2S]}

Multiply by [H3O+] both denominator and numerator,

Kc = {[NO2-][H3O+]/[ HNO2]} / {[ HS][H3O+] /[H2S]}

Just observe numerator and denominator,

We can write,

Kc = Ka (HNO2) / Ka (H2S)

Kc = (4.6x10-4)/(8.9x10-8)

Kc = 5.2 x 103

As Kc > 0 i.e. +ve reaction will be favored in specified direction i.e. product favored.

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2) At Kw = 2.00 x 10-13

Let us find pKw,

pKw = -log[Kw]

pKw = -log(2.00 x 10-13)

pKw = 12.70

pKw is defined as,

At neutral condition pH = pOH = pKw / 2

pH = 12.70/2

pH = 6.35

At pH < 6.35 ……… Solution acidic   and

pH > 6.35 ……. Solution will be basic.

At 75 0C pH = 6.78 > 6.35 and hence solution is BASIC.

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3)

               NH4Cl (aq. -----> NH4+ (aq.) + Cl- (aq.)

               0.245 M             0.245 M      0.245 M

                                 NH4+ (aq.) + H2O <------> NH3 + [H3O+]

Initial conc.              0.245 M                             0         0

Equ. Conc.               (0.245-X)                            X         X

Kb(NH3) = (1.76 x 10-5)

Ka x Kb = Kw = 1 x 10-14

Ka = (1x10-14)/Kb

Ka = (1x10-14)/(1.76x10-5)

Ka = 5.68 x 10-10

Ka for this ionization equilibrium is,

Ka = [NH3][ [H3O+]]/[NH4+]

Ka = (X)(X)/(0.245-X)

X <<< 0.245 hence 0.245-X = 0.245 (apprx.)

Ka = X2/0.245

X2 = 0.245 x (5.68 x 10-10)

X2 = 5.68 x 10-10

Taking square root of both sides,

X = 1.8 x 10-5 M

Hence,

[H3O+] = 1.8 x 10-5 M

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14)

H2SO3 is the weakest acid among all.

HI and HBr ionizes to (H+ and) I- and Br- which are stable conjugate bases.

HNO3 ionizes to conjugate Base NO3- is stabilized by strong conjuagation in N=O and has stabilization by strong electron withdrawer N and O atoms.

HClO4 ionizes to conjugate base ClO4- which is highly stabilized by conjugation in Cl=O and strong electron withdrawer Cl and 4 O atoms.

But in H2SO3 conjuagte base is HSO3- . In this conjugate base only one S=O group is there for conjugation and S is less electronegative that other central atoms in above cases.

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