University:
Glendale Community College
Course:
CHM130 | Fundamental Chemistry
Academic year:

2019
Views:

118

Pages:

2
Author:

craquetszs

Acid Base Key 1. Write the formula for the conjugate acid of the following bases: a. CN- conjugate acid is _____HCN____ b. HCO3c. NH3 d. PO43- conjugate acid is _________H2CO3___________ conjugate acid is ___ NH4+________ conjugate acid is _____ HPO42-__________ 2. Write the balanced reaction for what happens when hydrobromic acid is put in water. Draw the resulting solution in the beaker. HBr(aq) + H2O(l) Br - (aq) + H3O+(aq) 3. Write the balanced reaction for what happens when acetic acid is put in water. Draw the resulting solution in the beaker. CH3COOH (aq) + H2O(l) CH3COO- (aq) + H3O+(aq) 4. Write the balanced reaction for what happens when lithium hydroxide is put in water. Draw the resulting solution in the beaker. LiOH (s) Li+(aq) + OH-(aq) 5. If the pOH of a solution is 4.52, calculate the pH, the [H+] and the [OH-]. Is this solution acidic or basic? ____basic, more OH- than H+____ pH = 9.48 [H+] = 3.3 x 10-10 M [OH-] = 3.0 x 10-5M 6. Calculate the pH for a 1.55 M solution of pyridine, an amine, (C5H5N). Kb = 1.7 x 10-9 C5H5N(aq) + H2O(l) I 1.55 --C -x --E (1.55-x) --- C5H6N+(aq) + OH-(aq) 0 0 +x +x (x) (x) 1.7 x 10-9 = x2 / 1.55 (with approximation) x = 5.133 x 10-5 = [OH-] check approximation = good ☺ pOH = 4.29 and pH = 9.71 7. Calculate the concentration for a solution of hydroiodic acid that has a pH of 2.583. So 2.583 = -log[H+] thus [H+] = 2.61 x 10-3M Since all the HI acid ionizes, the original [HI] also = 2.61 x 10-3M 8. Identify the following as strong or weak acids, strong or weak bases, neutral salts, basic salts or acidic salts. HClO4 ___ strong acid ___, NH3____ weak base ______, NH4NO3___ acidic salt _____, H2SO4___ strong acid ____, Ba(OH)2____ strong base ____, LiCH3COO___ basic salt ____, HF_____ weak acid ___, NaF_______ basic salt ______, KOH_______ strong base ________, AlBr3____ acidic salt ____, K2CO3_____ basic salt ____, Ba(NO3)2____ neutral salt _______. 9. A 0.125M weak monoprotic acid solution has a pH of 4.25. What is the solution’s % ionization? % ion = [H+] / [acid initial] x 100 % ion = (5.6234 x 10-5M / .125) x 100 = .045 % ionized 10. Do this problem on the back of this page. Show all your work including the reactions. Calculate the pH if 3.33 grams of potassium acetate is dissolved in 3.50 liters of water. Ka for acetic acid is 1.8 x 10-5 3.33 g KCH3COO ( mol / 98.144g) (1 / 3.50L) = 9.6942 x 10-3 M KCH3COO is a soluble salt: KCH3COO basic K+(aq) + CH3COO-(aq) K+ ion is neutral, acetate is Because all the salt dissolves, the [acetate ion] = 9.6942 x 10-3 M CH3COO-(aq) + H2O(l) Kb) I .0096942 C -x E (0.0096942-x) ----- CH3COOH (aq) + OH-(aq) 0 x x Kb = Ka / Kw = (base in water needs 0 x x Kb = 5.5556 x 10-10 = x2 / 0.0096942 (approximation) then check approximation good ☺ -6 x = 2.3207 x 10 M = [OH ] ( IF this was the final answer it would have 2 sig cause Ka had 2 sig) pOH = 5.63 so pH = 8.37 Ka has 2 sig fig so pH has 2 decimal places