For the following acid-base reaction H₂S(aq) + CN⁻(aq) ⇌ HS⁻(aq) + HCN(aq) ∆H° = -24.7 kJ/mol...

For the following acid-base reaction H₂S(aq) + CN⁻(aq) ⇌ HS⁻(aq) + HCN(aq) ∆H° = -24.7 kJ/mol and ∆S° = -49.9 J/mol･K. If you mix 100 mL of 0.0150 M NaCN with 100 mL of 0.0150 M H₂S, after equilibrium is established at 25°C what will be the molar concentration of HCN?

Expert Solution

queen_honey_blossom answered 2 years ago

The enthalpy of reaction for HCN(aq) + OH-(aq) -> CN-(aq) + H2O(l) is -12.1 kJ/mol. Assuming...

The enthalpy of reaction for HCN(aq) + OH-(aq) -> CN-(aq) + H2O(l) is -12.1 kJ/mol. Assuming that the correct enthalpy of reaction of the experiment is -54.0 kJ/mol, calculate the enthalpy change for the ionization of HCN: HCN(aq) -> H+(aq) + CN-(aq) Be sure to report your answer to the correct number of significant figures. ΔH = ____ kJ/mol

a) identify the conjugate acid-base pairs for the following reaction: H2O (aq) + CN- (aq)--->HCN (aq)+OH-...

a) identify the conjugate acid-base pairs for the following reaction: H2O (aq) + CN- (aq)--->HCN (aq)+OH- (aq) b) give the conjugate acid of HSO4- and PO4 -2 c) give the conjugate base of HCLO4 and H3O+

Consider the following dissociation of the weak acid, HCN: HCN(aq) ⇌ CN−(aq) + H+(aq) K =...

Consider the following dissociation of the weak acid, HCN: HCN(aq) ⇌ CN−(aq) + H+(aq) K = 6.2×10-10 A solution is made with an initial concentration of 2.60 M HCN. At equilibrium, what is the concentration of H+ ions in solution? (Hint: You may use the 5% approximation.) ?M

Consider the acid-base equation HF(aq) + CN‾(aq) → (equilibrates with) F‾(aq) + HCN(aq). The Ka of...

Consider the acid-base equation HF(aq) + CN‾(aq) → (equilibrates with) F‾(aq) + HCN(aq). The Ka of HF is 7.1x10-4, while the Kb of CN‾ is 2.0x10-5. (a) Label each reactant and product as stronger acid, stronger base, weaker acid, or weaker base (b) Write out the Ka chemical equations for HF and for HCN. (c) Calculate Kc for the reaction HF(aq) + CN‾(aq) → F‾(aq) + HCN(aq)

identify acid, base, in the reaction HSO3- +CN-= HCN +SO3-2

Calculate the value of K in the following reaction: HCN (aq) + F- (aq) <-------> CN-...

Calculate the value of K in the following reaction: HCN (aq) + F- (aq) <-------> CN- (aq) + HF (aq) Ka for HCN = 6.2 x 10^-10 Ka for HF = 7.2 x 10^-4

The reaction of the weak acid HCN with the strong base KOH is: HCN(aq)+KOH(aq)-->HOH(l)+KCN(aq) To compute...

The reaction of the weak acid HCN with the strong base KOH is: HCN(aq)+KOH(aq)-->HOH(l)+KCN(aq) To compute the pH of the resulting solution if 54mL of 0.79M HCN is mixed with 2.0 × 10^1 mL of 0.32 KOH we need to start with the stoichiometry. Let\'s do just the stoich in steps:. a)How many moles of acid? b)How many moles of base? c)What is the limiting reactant? d)How many moles of the excess reagent after reaction? e)What is the concentration of...

Complete the following tables: Acid Conjugate Base Base Conjugate Acid H2O H3O+ CN- HS- HNO2 NH3...

Complete the following tables: Acid Conjugate Base Base Conjugate Acid H2O H3O+ CN- HS- HNO2 NH3 HSO4- HCO3-

Consider the reaction HCN(aq) H+(aq) +CN-(aq). the equilibrium constant is 6.2*10^-10. If you place 0.4 mols...

Consider the reaction HCN(aq) H+(aq) +CN-(aq). the equilibrium constant is 6.2*10^-10. If you place 0.4 mols of HCN in a 2.0 liter flask, what is the equilbrium of CN-? so far ive got to x=x(0.2)(6.2*10^-10)

Consider the dissociation of aqueous HCN at 25°C: HCN(aq) --> H+(aq) + CN–(aq) K = 6.2×10–10...

Consider the dissociation of aqueous HCN at 25°C: HCN(aq) --> H+(aq) + CN–(aq) K = 6.2×10–10 . a) Compute ΔG° at 25°C. b) If 0.120 mol of HCN is dissolved to make 200. mL of solution, then what are the equilibrium concentrations of all of the species? c) Suppose 0.040 mol of HCN is dissolved in the same solution without appreciably increasing the volume. Compute the derivative dG/dξ for the system before it has a chance to re-establish equilibrium. Use...

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