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- Write the expression for K, and K, for the reaction PH;BCl,(s) = PH;(e) + BCl(g) 1.Propionic acid, HC3H5O2, has Ka= 1.34 x 10–5. (a) What is the molar concentration of H3O+ in 0.15 M HC3H5O2 and the pH of the solution? (b) What is the Kb value for the propionate ion, C3H5O2–? (c) Calculate the pH of 0.15 M solution of sodium propionate, NaC3H5O2. (d) Calculate the pH of solution that contains 0.12 M HC3H5O2 and 0.25 M NaC3H5O2.The pOH of an aqueous solution of 0.588 M benzoic acid, C₂H5COOH is
- COHSOH(ag) + H2On + CeHsO (aq) + H3O*(a9) Ka= 1.12 x 10-10 (a) Phenol is a weak acid that partially dissociates in water according to the equation above. Write the equilibrium-constant expression for the dissociation of the acid in water. (b) What is the pH of a 0.75 M CaHsOH(ag) solution? (C) For a certain reaction involving CaHsOH(ag) to proceed at a significant rate, the phenol must be primarily in its deprotonated form, C3H5O (eg). In order to ensure that the CsHsOH(aq) is deprotonated, the reaction must be conducted in a buffered solution. On the number scale below, circle each pH for which more than 50 percent of the phenol molecules are in the deprotonated form (CoHsO (aq). Justify your answer. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Justification: (d) CeHsOH(ag) reacts with NaOH(ag). Write a net ionic equation representing this reaction (aka: invasion equation). (e) What is the pH of the resulting solution when 30 mL of 0.40 M CSH5OH(aq) is added to 25 mL of 0.60 M NAOH. Show all work…A weak acid, HA, is a monoprotic acid. A solution that is 0.140 Min HA has a pH of 1.800 at 25°C. HA(aq) + H,0(1) 2 H;0"(aq) + A (aq) What is the acid-ionization constant, K, for this acid? What is the degree of ionization of the acid in this solution? Ka- Degree of ionization =(a) Calculate the OH¯ concentration in an aqueous solution at 25°C with an H3O™ concentration of 1.03 x 10 M. х 10 M (b) The value of K, at 50°C is 5.48 × 10¬14. Calculate the OH¯ concentration from the above solution at 50°C. x 10
- CH3CH2CO2H , Ką = 1.3 x 105 CH3CH2CO2 , Kp = 7.7 x 10-10 F, Kp = 1.4 x 10-11 HF , Ką = 7.2 x 10-4 For the reaction between propanoic acid (CH3CH2CO2H) and a solution containing fluoride ions (F"): (a) The name of the conjugate base produced is (b) The name of the conjugate acid produced is (c) The reaction is favoured (write 'reactant' or 'product'), because the reactants are the (write 'weaker' or 'stronger') set of acid and base.Calculate the pH and the pOH of these solutions: (c) 0.957 M HC2H3O2 (Ka = 1.8 x 10-5) pH = pOH =2. Determine [H3O+] and [OH] concentrations (in M to two decimal places) for each: (a) Pure water at 10 °C (K = 0.293 x 10-14) W (b) An aqueous solution with a pH of 3.23 A: 5.41x108; 5.41×10-8 A: 5.89x104; 1.70×10-11
- Acetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions: HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!). (d) When starting with completely un-dissociated acetic acid, is it accurate to assume that [HOAc]0 = [HOAc]eq? Why or why not? (e) A highly concentrated acetic acid solution contains 15.0M acetic acid at equilibrium. What are the equilibrium concentrations of the hydronium and acetate ions in this solution? (f) Creating the concentrated acetic acid solution by dissolving liquid HOAc in water raises the temperature of the water by about 5°C from room temperature. At 50°C, do you expect the solution to contain more or less acetate ion OAc– than what you calculated in (c)? Why?Acetic acid is a weak acid, meaning it does not fully dissociate in water. Instead, there is an equilibrium between the dissolved but undissociated molecule and the component ions: HOAc (aq) + H2O (l) ⇌ H3O+ (aq) + OAc– (aq)OAc– is an abbreviation for the acetate ion, CH3COO–, and H3O+ is the hydronium ion (lone protons, H+ (aq), do not exist!). (a) Write the equilibrium constant expression for the dissociation of acetic acid. (b) Vinegar sold commercially is typically 0.8 − 1.0 M acetic acid. A 1.00 M solution of acetic acid is measured by its pH to have an equilibrium concentration of 4.19×10−3 M for both acetate ions and hydronium ions at room temperature. Assuming [HOAc]0 = 1.00M, what is the equilibrium concentration of undissociated acetic acid [HOAc]eq to the correct number of significant figures? (c) What is the value of the equilibrium constant Keq for the dissociation according to the concentrations from part (b)? (d) When starting with completely un-dissociated…Rank the following aqueous solutions in terms of pH. Why Are they in that order? (a)0.1 M CH3COOH (b)0.1 M H2SO4 (c)0.1 M HNO3 (d)0.1 M CH3COOH + 0.1 M CH3COONa (e)0.1 M HF (f)0.1 M HCl
