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Worksheet: Acids, Bases & pH Key. 1. Label the substance below as an acid, base or neutral substance. pH = 7 (b) pOH = 9 (c) pH = 3 (d) pOH = 4 (e) pH = 13 (f) pOH = 7. 1. Label the substance below as an acid, base or neutral substance. pH = 7 N (b) pOH = 9 A
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1. Label the substance below as an acid, base or neutral substance. • pH = 7 (b) pOH = 9 (c) pH = 3 (d) pOH = 4 (e) pH = 13 (f) pOH = 7
1. Label the substance below as an acid, base or neutral substance. • pH = 7 N (b) pOH = 9 A (c) pH = 3 A (d) pOH = 4 B (e) pH = 13 B (f) pOH = 7 N
2. How does the Arrhenius theory define an acid? • An acid produces H3O+ ions when dissolved in water.
3. How does the Arrhenius theory define an base? • A base produces OH- ions when dissolved in water.
4. Differentiate between an electrolyte and a nonelectrolyte.
4. Differentiate between an electrolyte and a nonelectrolyte. • An electrolyte dissolves into ions when placed in water whereas a nonelectrolyte dissolves into molecules.
5. What type of substances are electrolytes? Acids, Bases, Salts
6. When H+ ions are dissolved in water they become hydronium ions. What is the formula for a hydronium ion?
6. When H+ ions are dissolved in water they become hydronium ions. What is the formula for a hydronium ion? H3O+
7. What is the pH of a solution with a [H+] = 4.32 x 10-12? pOH = 11.365
8. What is the pOH of a solution with a [OH-] = 6.83 x 10-3?
8. What is the pOH of a solution with a [OH-] = 6.83 x 10-3? pOH = 2.166
9. What is the pOH of a solution with a pH = 3.8? pOH = 10.2
10. What is the [H+] in a solution with a pH = 7.3? [H+] = 5.0 x 10-8
11. What is the [OH-] in a solution with a pOH = 10.7? [OH-] = 2.0 x 10-11
12. What is the pH of a solution with a [OH-] = 8.62 x 10-5?
12. What is the pH of a solution with a [OH-] = 8.62 x 10-5? pH = 9.936
13a) Write the balanced equation for the acid-base neutralization of H2SO4 and Ba(OH)2.
13a) Write the balanced equation for the acid-base neutralization of H2SO4 and Ba(OH)2. H2SO4 + Ba(OH)2 → BaSO4 + 2H2O
13b) Write the balanced equation for the acid-base neutralization of HF and NaOH.
13b) Write the balanced equation for the acid-base neutralization of HF and NaOH. HF + NaOH→ NaF + H2O
13c) Write the balanced equation for the acid-base neutralization of HI and Sr(OH)2.
13c) Write the balanced equation for the acid-base neutralization of HI and Sr(OH)2. 2HI + Sr(OH)2 → SrI2 + 2H2O
13d) We often use the equation Macid x Vacid = Mbase x Vbase. Which of the acid-base neutralization reactions above could we not use this formula for? Explain.
13d) We often use the equation Macid x Vacid = Mbase x Vbase. Which of the acid-base neutralization reactions above could we not use this formula for? Explain. • The reaction between HI and Sr(OH)2 could not be used because the acid and the base do not react in a 1:1 ratio.
13e) In an acid-base neutralization reaction 26 ml of Ba(OH)2 reacts with 16 ml of 0.48M H2SO4. What is the molarity of the Ba(OH)2?
13e) In an acid-base neutralization reaction 26 ml of Ba(OH)2 reacts with 16 ml of 0.48M H2SO4. What is the molarity of the Ba(OH)2? 0.30M
13f) How many milliliters of 0.57M HF will react with 39 ml of 0.86M NaOH in an acid-base neutralization reaction?
13f) How many milliliters of 0.57M HF will react with 39 ml of 0.86M NaOH in an acid-base neutralization reaction? 59 ml