Lesson 15: Acid-Base Stoichiometry
A neutralization reaction occurs between a strong acid and a strong base.
The two products are always water (formed by the H+ ion joining with the OH- ion) and an ionic compound. Sometimes the ionic compound is called a salt.
eg. Ba(OH)2(aq) + HCl(aq) 🡪 H2O(l) + BaCl2(aq)
Now, balance the equation:
___ Ba(OH)2(aq) + ___ HCl(aq) 🡪 ___ H2O(l) + ___ BaCl2(aq)
Hint: To balance a neutralization reaction, first balance the # of H+ and OH- ions. They always have to be the same # to make water.
Ba(OH)2(aq) + 2HCl(aq) 🡪 2H2O(l) + BaCl2(aq)
H2SO4(aq) + 2KOH(aq) 🡪 2H2O(l) + K2SO4(aq)
0.0125 L 0.100L
2.0 mol/L 0.50 mol/L
0.025 mol 0.05 mol
1 2
0.025 0.025
A 10 mL sample of NaOH of an unknown concentration is titrated with 0.245 M sulfuric acid. The results are summarized below. Fill in the chart below and use the reaction below to calculate the concentration of NaOH.
2NaOH(aq) + H2SO4(aq) 🡪 Na2SO4(aq) + 2H2O(l)
trial 1 | trial 2 | trial 3 | |
---|---|---|---|
Final reading | 15.7 mL | 30.6 | 45.7 |
Initial reading | 0.2 mL | 15.7 | 30.6 |
vol of H2SO4 added | 15.5 mL | 14.9 | 15.1 |
Find the average of the 3 trials
(15.5 + 14.9 + 15.1)/3 = 15.17
Now do the neutralization stoichiometry
2NaOH(aq) + H2SO4(aq) 🡪 Na2SO4(aq) + 2H2O(l)
Ba(OH)2(aq) + 2HCl(aq) 🡪 2H2O(l) + BaCl2(aq)
V (L) | 0.0145 | 0.050 |
---|---|---|
C (mol/L) | 0.352 | 0.204 |
n (mol) | 0.0051 | 0.0102 |
coeff | 1 | 2 |
WMR | 0.0051 | 0.0051 |
Mg(OH)2(s) + 2HCl(aq) 🡪 2H2O(l) + MgCl2(aq)
m (g) | 0.32 | 0.0171 | V (L) |
---|---|---|---|
M (g/mol) | 58.18 | 0.643 | C (mol/L) |
n (mol) | 0.00550 | 0.0110 | n (mol) |
coeff | 1 | 2 | coeff |
WMR | 0.0055 | 0.0055 | WMR |
A neutralization reaction occurs between a strong acid and a strong base.
The two products are always water (formed by the H+ ion joining with the OH- ion) and an ionic compound. Sometimes the ionic compound is called a salt.
eg. Ba(OH)2(aq) + HCl(aq) 🡪 H2O(l) + BaCl2(aq)
Now, balance the equation:
___ Ba(OH)2(aq) + ___ HCl(aq) 🡪 ___ H2O(l) + ___ BaCl2(aq)
Hint: To balance a neutralization reaction, first balance the # of H+ and OH- ions. They always have to be the same # to make water.
Ba(OH)2(aq) + 2HCl(aq) 🡪 2H2O(l) + BaCl2(aq)
H2SO4(aq) + 2KOH(aq) 🡪 2H2O(l) + K2SO4(aq)
0.0125 L 0.100L
2.0 mol/L 0.50 mol/L
0.025 mol 0.05 mol
1 2
0.025 0.025
A 10 mL sample of NaOH of an unknown concentration is titrated with 0.245 M sulfuric acid. The results are summarized below. Fill in the chart below and use the reaction below to calculate the concentration of NaOH.
2NaOH(aq) + H2SO4(aq) 🡪 Na2SO4(aq) + 2H2O(l)
trial 1 | trial 2 | trial 3 | |
---|---|---|---|
Final reading | 15.7 mL | 30.6 | 45.7 |
Initial reading | 0.2 mL | 15.7 | 30.6 |
vol of H2SO4 added | 15.5 mL | 14.9 | 15.1 |
Find the average of the 3 trials
(15.5 + 14.9 + 15.1)/3 = 15.17
Now do the neutralization stoichiometry
2NaOH(aq) + H2SO4(aq) 🡪 Na2SO4(aq) + 2H2O(l)
Ba(OH)2(aq) + 2HCl(aq) 🡪 2H2O(l) + BaCl2(aq)
V (L) | 0.0145 | 0.050 |
---|---|---|
C (mol/L) | 0.352 | 0.204 |
n (mol) | 0.0051 | 0.0102 |
coeff | 1 | 2 |
WMR | 0.0051 | 0.0051 |
Mg(OH)2(s) + 2HCl(aq) 🡪 2H2O(l) + MgCl2(aq)
m (g) | 0.32 | 0.0171 | V (L) |
---|---|---|---|
M (g/mol) | 58.18 | 0.643 | C (mol/L) |
n (mol) | 0.00550 | 0.0110 | n (mol) |
coeff | 1 | 2 | coeff |
WMR | 0.0055 | 0.0055 | WMR |