Multiple ChoiceCalculate the pH of a 0.03 M aqueous solution of a weak acid, HA, that has a pKa of 3.39.
Multiple ChoiceWhat is the pH of a 0.10 M solution of hypochlorous acid (HOCl) given that the acid dissociation constant (Ka) is 3.0 x 10^-8?
Multiple ChoiceCalculate the pH of a 0.300 M NaHCOO solution, given that the Ka for HCOOH is 1.8 x 10^-4.
Multiple ChoiceCalculate the pH of a 0.800 M NaCH3CO2 solution. Ka for acetic acid, CH3CO2H, is 1.8 × 10^-5.
Multiple ChoiceCalculate the pH of a buffer that contains 1.5 M CH3COOH and 1.3 M NaCH3COO. Given that the acid dissociation constant (Ka) for CH3COOH is 1.8 x 10^-5.
Multiple ChoiceWhat is the percent ionization of a 0.16 M benzoic acid solution in a solution containing 0.28 M sodium benzoate, given that the Ka of benzoic acid is 6.3 x 10^-5?
Multiple ChoiceConsider the titration of 25.00 mL of 0.150 M hydrazoic acid (Ka = 1.90 x 10^-5) with 0.200 M NaOH. (Hydrazoic acid is HN3, not to be confused with ammonia, NH3) What is the pH of the hydrazoic acid solution prior to beginning the titration?
Multiple ChoiceWhat is the pH of a 0.1 M solution of acetic acid (CH3COOH) given that its acid dissociation constant (Ka) is 1.8 x 10^-5?
Multiple ChoiceDetermine the pH of a 0.036 M solution of formic acid (HCO2H), given that the acid dissociation constant (Ka) is 1.8 x 10^-4. The dissociation reaction is: HCO2H (aq) ⇌ H+ (aq) + HCO2- (aq).
Multiple ChoiceWhat is the pH of a 0.461 M benzoic acid (C6H5CO2H) solution if the acid dissociation constant (Ka) is 6.5 x 10^-5?
Multiple ChoiceDetermine the pH of a buffer solution prepared by dissolving 0.20 mol of formic acid (HCHO) and 0.80 mol of sodium formate (NaCHO) in enough water to make 1.0 L of solution. The pKa of formic acid is 3.75.
Multiple ChoiceWhat is the pH of a 0.10 M solution of H₂SO₃, given that its pKₐ is 1.77 and the change in concentration is small?
Multiple ChoiceWhat is the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of 0.1249 M NaOH? (Ka for aniline = 2.40 x 10^-5)