Chemistry: Behavior of Gases - Unit 5
Name:_________KEY___________Hour:____Date:___________Chemistry: Partial Neutralization Directions: Solve the following problems. Show your work and include units on your answers.1. 7.35 L of 0.411 M HI were mixed with 10.26 L of 0.289 M NaOH. Find the pH of the resulting solution. 0.05571 mol H + excess mol H + = 0.411 M (7.35 L) = 3.02085 mol H +mol OH –= 0.289 M (10.26 L) = 2.96514 mol OH –2. 68.1 mL of 2.75 M HNO3 were mixed with 61.0 mL of 3.11 M KOH. Find the pH of the resulting solution. 0.002435 mol OH – excess mol H + = 2.75 M (0.0681 L) = 0.187275 mol H +mol OH –= 3.11 M (0.0610 L) = 0.18971 mol OH –pH = 12.283. 13.8 L of 4.28 M sulfuric acid were mixed with 40.1 L of 2.96 M cesium hydroxide. Find the pH of the resulting solution. 0.568 mol OH – excess mol H + = 4.28 M (2) (13.8 L) = 118.128 mol H +mol OH –= 2.96 M (40.1 L) = 118.696 mol OH –pH = 12.024. 27.2 L of 3.81 M hydrobromic acid were mixed with 72.8 L of sodium hydroxide to produce a solution having a pH of 3.11. What was the concentration of the base?Find mol of excess H + = 7.7625 x 10–4 M (27.2 L + 72.8 L) = 7.7625 x 10–2 mol H +So now… 3.81 M H + (27.2 L) – [OH –] (72.8 L) = 7.7625 x 10–2 mol H +Solve for… [OH –] = [NaOH] = 1.42 MANSWERS:1. 2.502. 12.283. 12.024. 1.42 M ................
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