Download presentation
Presentation is loading. Please wait.
1
Chapter 8 Acids and Bases
8.5 Reactions of Acids and Bases
2
Acids and Metals Acids react with metals
such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn to produce hydrogen gas and the salt of the metal Molecular equations: 2K(s) HCl(aq) 2KCl(aq) + H2(g) metal acid salt hydrogen gas Zn(s) HCl(aq) ZnCl2(aq) + H2(g) metal acid salt hydrogen gas
3
Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Label the metal, acid, and salt.
4
Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Mg(s) HCl(aq) MgCl2(aq) + H2(g) metal acid salt
5
Acids and Carbonates Acids react
with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water 2HCl(aq) + CaCO3(s) CO2(g) + CaCl2(aq) + H2O(l) acid carbonate carbon salt water dioxide HCl(aq) + NaHCO3(s) CO2(g) + NaCl (aq) + H2O(l) acid bicarbonate carbon salt water
6
Learning Check Write a balanced equation for the following reactions:
A. MgCO3(s) + HBr(aq) B. HCl(aq) + NaHCO3(aq)
7
Solution Write a balanced equation for the following reactions:
A. MgCO3(s) + 2HBr(aq) MgBr2(aq) + CO2(g) + H2O(l) carbonate acid salt carbon water dioxide B. HCl(aq) + NaHCO3(aq) NaCl(aq) + CO2(g) + H2O(l) acid bicarbonate salt carbon water
8
Neutralization Reactions
In a neutralization reaction, an acid reacts with a base to produce salt and water the acid HCl reacts with NaOH to produce salt and water the salt formed is the anion from the acid and cation of the base HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) acid base salt water
9
Neutralization Reactions
In neutralization reactions, if we write the strong acid and strong base as ions, we see that H+ reacts with OH− to form water, leaving the ions Na+ and Cl in solution: H+(aq) + Cl(aq) + Na+(aq) + OH(aq) Na+(aq) + Cl(aq) + H2O(l) the overall reaction is H3O+ from the acid and OH from the base form water: H+(aq) OH(aq) H2O(l)
10
Guide for Balancing Neutralization Reactions
11
Balancing Neutralization Reactions
Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. Step 1 Write the reactants and products. Mg(OH)2 + HNO3 Step 2 Balance the H+ in the acid with the OH in the base. Mg(OH) HNO3 Step 3 Balance the H2O with H+ and the OH. Mg(OH) HNO3 salt + 2H2O Step 4 Write the salt from the remaining ions. Mg(OH) HNO3 Mg(NO3) H2O
12
Learning Check Select the correct group of coefficients for each of the following neutralization equations. 1. HCl(aq) + Al(OH)3(aq) AlCl3(aq) + H2O(l) A. 1, 3, 3, B. 3, 1, 1, C. 3, 1, 1, 3 2. Ba(OH)2(aq) + H3PO4(aq) Ba3(PO4)2(s) + H2O(l) A. 3, 2, 2, B. 3, 2, 1, C. 2, 3, 1, 6
13
Solution 1. HCl(aq) + Al(OH)3(aq) AlCl3(aq) + H2O(l)
Step 1 Write the reactants and products. HCl + Al(OH)3 Step 2 Balance the H+ in the acid with the OH in the base. 3HCl + Al(OH)3 Step 3 Balance the H2O with H+ and the OH. 3HCl + Al(OH)3 salt + 3H2O Step 4 Write the salt from the remaining ions. 3HCl(aq) + Al(OH)3(aq) AlCl3(aq) + 3H2O(l) The answer is C. 3, 1, 1, 3.
14
Solution 2. Ba(OH)2(aq) + H3PO4(aq) Ba3(PO4)2(s) + H2O(l)
Step 1 Write the reactants and products. Ba(OH)2 + H3PO4 Step 2 Balance the H+ in the acid with the OH in the base. 3Ba(OH)2 + 2H3PO4 Step 3 Balance the H2O with H+ and the OH. 3Ba(OH)2 + 2H3PO4 salt + 6H2O Step 4 Write the salt from the remaining ions. 3Ba(OH)2(aq) + 2H3PO4(aq) Ba3(PO4)2(s) + 6H2O(l) The answer is B. 3, 2, 1, 6.
15
Basic Compounds in Antacids
16
Learning Check Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: Al(OH)3 and Mg(OH)2
17
Solution Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta. Mylanta: For Al(OH)3: Step 1 Write the reactants and products. Al(OH)3 + HCl Step 2 Balance the H+ in the acid with the OH in the base. Al(OH)3 + 3HCl Step 3 Balance the H2O with H+ and the OH. Al(OH)3 + 3HCl salt + 3H2O Step 4 Write the salt from the remaining ions. Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)
18
Solution Write the neutralization reactions for stomach acid, HCl, and
the ingredients in Mylanta. Mylanta: For Mg(OH)2: Step 1 Write the reactants and products. Mg(OH)2 + HCl Step 2 Balance the H+ in the acid with the OH in the base. Mg(OH)2 + 2HCl Step 3 Balance the H2O with H+ and the OH. Mg(OH)2 + 2HCl salt + 2H2O Step 4 Write the salt from the remaining ions. 2HCl(aq) + Mg(OH)2(aq) MgCl2(aq) + 2H2O(l)
19
Acid–Base Titration Titration
is a laboratory procedure used to determine the molarity of an acid uses a base such as NaOH to neutralize a measured volume of an acid Base NaOH Acid Solution
20
Indicator An indicator is added to the acid in the flask
causes the solution to change color when the acid is neutralized
21
Endpoint of Titration At the endpoint,
the indicator gives the solution a permanent pink color the volume of the base used to reach the endpoint is measured the molarity of the acid is calculated using the neutralization equation for the reaction
22
Guide to Calculations for Acid–Base Titrations
23
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Step 1 State given and needed quantities. Given: 18.5 mL of M NaOH 10.0 mL of HCl Needed: Molarity of HCl
24
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Step 2 Write a plan to calculate molarity or volume. molarity equation molarity NaOH coefficients HCl 18.5 mL L moles NaOH moles HCl L HCl
25
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Step 3 State equalities and conversion factors including concentration. 1 L = 1000 mL 0.225 mole NaOH and mole HCl x 1 L NaOH mole NaOH
26
Acid–Base Titration Calculations
What is the molarity of an HCl solution if 18.5 mL of M NaOH are required to neutralize 10.0 mL of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Step 4 Set up the problem to calculate the needed quantity mL NaOH 1 L NaOH mole NaOH 1000 mL NaOH 1 L NaOH 1 mole HCl = mole of HCl 1 mole NaOH MHCl = mole HCl = M HCl L HCl
27
Learning Check Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) A mL B mL C mL
28
Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) Step 1 State given and needed quantities. Given: M H2SO4 50.0 mL of 1.00 M KOH Needed: mL of H2SO4
29
Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) Step 2 Write a plan to calculate molarity or volume. molarity equation molarity KOH coefficients H2SO4 50.0 mL L moles KOH moles H2SO4 L H2SO4
30
Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) Step 3 State equalities and conversion factors including concentration. 1 L = 1000 mL 1 L H2SO and mole H2SO4 2.00 mole H2SO mole KOH
31
Solution Calculate the mL of 2.00 M H2SO4 required to neutralize mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) Step 4 Set up the problem to calculate the needed quantity. 50.0 mL 1 L = L 103 mL L KOH 1.00 mole KOH 1 mole H2SO4 1 L KOH mole KOH 1 L H2SO mL = 12.5 mL. 2.00 mole H2SO L H2SO4 The answer is A.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.