Determine the equilibrium constant for the reaction NH4 + (aq) + OH– (aq) ⇌ H2O(ℓ) + NH3 (aq) given that the Ka of NH4+ is 5.68×10–10 .
Ka of NH4+ is:
NH4+ (aq) <—> NH3(aq) + H+(aq)
So,
Ka = [NH3][H+]/[NH4+]
= 5.68*10^-10
The dissociation of water:
H2O(l) <—> H+(aq) + OH-(aq)
Kw =[H+][OH-]/H2O
= 1.0*10^-14
The given equation is:
NH4+(aq) + OH-(aq) <—> H2O(l) + NH3(aq)
Kc = [H2O][NH3]/[NH4+][OH-]
= Ka/Kw
= (5.68*10^-10)/ (1.0*10^-14)
= 5.68*10^4
Answer: 5.68*10^4
Determine the equilibrium constant for the reaction NH4 + (aq) + OH– (aq) ⇌ H2O(ℓ) +...
Calculate the concentration of OH- at equilibrium in the following reaction: NH3 (aq) + H2O (l) -> NH4+ (aq) + OH- (aq) (Kb = 1.8 x10^(-5)
Given the equilibrium constants for the equilibria, 2 NH4+(aq) + 2 H2O(ª) 2 NH3(aq) + 2 H3O+(aq); Kc = CH3COOH(aq) + H2O(ª) CH3COO−(aq) + H3O+(aq); Kc = determine Kc for the following equilibrium. CH3COOH(aq) + NH3(aq) CH3COO−(aq) + NH4+(aq) a. 3.08 × 104 b. 9.96 × 10-15 c. 3.25 × 10-5 d. 1.75 × 10-5 e. 1.00 × 1014
For the reaction NH3(aq) + H+ (aq) ⇋ NH4 + (aq), at 20°C the equilibrium concentrations were as follows: [NH3] = 2 x 10-4 M; [H+ ] = 2 x 10-4 M; and [NH4 + ] = 18 M. Calculate the equilibrium constant for the reaction.
Calculate the value for the equilibrium constant of this reaction: NH4+(aq)+HCO3-(aq)↔NH3(aq)+H2CO3(aq) where [NH4+]=0.25 M, [HCO3-]=0.42 M, [NH3]=0.0078 M, and [H2CO3]=0.018 M. answer= 1.3 *10^-3
For aq. solutions of salt NH4NO2, following reactions possible: NH4+ + NO2- -> NH3 + HNO2 k1? NH4+ + H2O -> H3O+ + NH3. ka = 5.6 x 10^-10 NO2- + H2O -> HNO2 + OH- kb =2.2 x 10^-11 2H2O -> H3O+ + OH- kw = 1.0 x10^-14 Write symbolic expression for equilibrium constants for each reaction. Derive expression for k1 in terms of ka, kb, and kw; find numerical value of k1.
Reaction . NH3(aq) + H2O (l) D NH4+1(aq) + OH-1(aq) 16. Another student says “Adding MgCl2 cannot affect this reaction, because it doesn’t contain any MgCl2 as a reactant or product.” Choose the best answer below (Circle, bold, highlight, italicize one). The other student is correct. Only chemicals that are contained in the given chemical reaction affect reaction equilibrium. According to Le Chatelier’s principle, the addition of MgCl2 will not cause the reaction to shift left or right. The other...
Given the thermodynamic data below, determine the equilibrium constant for the following two reactions at 25 oC. Predict what will happen to ΔG, how the direction of the reaction will shift, and how the spontaneity will be affected when more BaSO4(s) is added to reaction a), and more NH4+(aq) is added to reaction b). Substance ∆Gof(kJ/mol) Substance ∆Gof(kJ/mol) Substance ∆Gof kJ/mol) BaSO4(s) –1353.1 SO42–(aq) –741.99 NH3(aq) –26.5 Ba2+(aq) –560.66 OH–(aq) –157.30 H2O(l) –237.2 NH4+(aq) –79.5 a. BaSO4(s) ⇌ Ba2+(aq) +...
Please help Is H2PO4- + H2O → H3O+ + HPO42- an acid or base? Is H2O(ℓ) + NH3(aq) -> NH4+ + OH- an acid or base Is HSO4- + H2O → OH- + H2SO4 an acid or base Is O2- + H2O → OH- + OH- an acid or base Is H3O+ + Cl- → H2O + HCl an acid or base
The equlibrium constant for the equation NH3(aq) + H2O(l) ↔ NH4+(aq) + OH⁻(aq) is Kb= 1.8×10-5 at 25 ºC. Calculate ΔGºrxn at 25 ºC. a) -2.27 kj/mol b) 0 kj/mol c) 2.27 kj/mol d) 267 J/mol 27.1 kj/mol
onsider the reaction of NH3 with water: NH3 + H2O ↔ NH4+ + OH- (Kb=Kw/Ka) and also NH4+ + H2O ↔ NH3 + H3O+ (Ka = 5.70×10-10) be sure to show all work including the ICE Tables. a) Find the pH of 50.00 mL of 0.200M NH3 after the addition of 45.00 mL of 0.200M HCl. b) Find the pH of 50.00 mL of 0.200M NH3 after the addition of 50.00 mL of 0.200M HCl. c) Find the pH of...