Ch 14 - This assignment has all of the answers including some show your work questions PDF

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This assignment has all of the answers including some show your work questions with work shown....

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Aundreauna Womack Chapter 14 8/19/19 18.) Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: (a) 0.200 M HCl Since HCl and H+ are in 1:1 mole ration 0.200 M HCl give 0.200 M H+ ions pH = -log[H+] pH = -log[0.200] pH = 0.699 pH + pOH = 14 pOH = 14 - pH = 14-0.699 pOH = 13.3 (b) 0.0143 M NaOH NaOH and OH- are 1:1 mole ratio 0.0143 M NaOH gives 0.0143 M OH- ions pOH = -log[OH-] pOH = -log(0.0143) pOH = 1.85 pH + pOH = 14 pH = 14 - pOH pH = 14 - 1.85 pH = 12.15

(c) 3.0 M HNO3 HNO3(aq) --------> NO3-(aq) + H+(aq) HNO3: H+ = 1:1 3.0 M HNO3 gives 3.0 M H+

pH = -log[H+] pH = - log [3.0] pH =-0.477 pOH= 14 - pH pOH = 14 - (-0.477) = 14.5

(d) 0.0031 M Ca(OH)2 Ca(OH)2(aq) ------------> Ca+2(aq) + 2OH-(aq) Ca(OH)2 :OH- are in 1:2 mole ratio Concentration of [OH-] = 2MOH 0.031/Ca(OH)2 × DỊCa(OH)2 pOH = -log[OH-] = -log[0.062] pOH = 1.2 pH = 14-pOH pH = 14-1.2 pH =12.8

36.) What is the ionization constant at 25 °C for the weak acid (CH3)2 NH2 +, the conjugate acid of the weak base (CH3)2NH, Kb = 7.4 × 10−4? As we know that ka.kb= 10*-14 Ka= 10*-14/kb Ka= 10*-14/7.4x10*-4 Ka= 1.35x10*-11 Ionization constant ka= 1.35x10*-11 52.) Which of the following will increase the percent of HF that is converted to the fluoride ion in water? (a) addition of NaOH (b) addition of HCl (c) addition of NaF 60.) Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid: (a) F−

F- + H2O ---------------> HF + OHKb = 1.0 x 10^-14 / 6.3 x 10^-4 Kb = 1.59 x 10^-7 (b) NH4 + NH4+ ---------------> NH3 + H+ Ka = Kw / Kb Ka = 1.0 x 10^-14 / 1.8 x 10^-6 Ka = 5.56 x 10^-10 (c) AsO4 3− ) AsO4-3 + H2O ---------------> HAsO4-2 + OHKb1 = Kw / Ka3 Kb1 = 1.0 x 10^-14 / 3.2 x 10^-12 Kb1 = 3.12 x 10^-3 (d) (CH3)2NH2+ Ka = 1.85 x 10^-11 (e) NO2− Kb = 1.79 x 10^-11 (f) HC2 O4 – (as a base) Kb1 = 1.79 x 10^-13 78) Determine whether aqueous solutions of the following salts are acidic, basic, or neutral: (a) Al(NO3)3 Al(NO3)3 is an acid because it will accept the lone pair of electrons from the water molecule. (b) RbI RbI will simply dissociate in the form of ions they can't show any acidic or basic properties. (c) KHCO2 In aqueous solution it will give K+ and HCO2- ion and HCO2- ion is well known for basic properties. (d) CH3NH3Br It is an ammonium salt which on dissolving in water will give CH3NH3+ and Br- ions. In these two ions, CH3NH3+ can accept the electron from any base so it will acidic in nature.

90.) What is [H3O+] in a solution of 0.075 M HNO2 and 0.030 M NaNO2? HNO2(aq) + H2 O(l) ⇌ H3 O+ (aq) + NO2 −(aq) Ka = 4.5 × 10−5 concentration of hydrogenium ions is 9,54·10⁻⁵ M. c(HNO₂) = 0,075 M. c(NaNO₂) = 0,035 M. Ka(HNO₂) = 4,5·10⁻⁵. This is buffer solution, so use Henderson–Hasselbalch equation: pH = pKa + log(c(NaNO₂) ÷ c(HNO₂)). pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M). pH = 4,35 - 0,33. pH = 4,02. [H₃O⁺] = 10∧(-4,02). [H₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M. 98.) Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH2PO4, and enough water to make 0.500 L of solution.

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