Chemistry 2 AP/AC
Unit 2 HW 3
Name_________________________
1) Predict the products for the following reactions, then write the balanced, net ionic
reaction for each.
(a) Zn(s) + HCl(aq) →
(b) MgCO3(s) + HNO3(aq) →
(c) HC2H3O2(aq) + NaHCO3(aq) →
2) For the following neutralization reactions, determine the products and write the
balanced net ionic equations.
Mg(OH)2(s) 
(a)
HClO4(aq)
+
(b)
KOH(aq)
(c)
HC2H3O2(aq) +
(d)
H2CO3(aq)
(e)
HNO3(aq)
(f)
HF(aq) + NaOH(aq) →
(g)
NaNO2(aq) + HCl(aq) →
+
H3PO4(aq)
+
+
KOH(aq)


NaOH(aq)

Ba(OH)2(s)

3) Consider the titration of HCl with NaOH. How many mL of a 0.1105 M NaOH
solution would be required to neutralize 10.00 mL of a 0.8035 M solution of the acid?
4) 37.6 mL of a NaOH solution neutralizes 1.004 g of KHP (molar mass 204.23 g).
10.00 mL of an HCl solution is neutralized by 22.6 mL of the NaOH solution. What
is the molarity of the NaOH and HCl solutions?
5) A 2.20 gram sample of a monoprotic acid with an empirical formula of C3H4O3 is
dissolved in 1.00 L of water. A titration required 25.0 mL of 0.500 M NaOH to react
completely with all the acid present. Determine the molar mass and molecular
formula of the acid.
6) Carminic acid, a naturally occurring red pigment extracted from the cochineal insect,
contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the
first half of the nineteenth century. It is 53.66% C and 4.10 % H by mass. A
titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 grams of the
acid. Assuming the acid is monoprotic, determine the molecular formula of
carminic acid.
7) A particular acid containing just carbon, hydrogen and oxygen is known to be
triprotic (it has three protons that are given up during a titration.)
In one experiment, a 0.250 gram sample of the acid dissolved in 25.0 mL of water
requires 37.2 mL of 0.105 M NaOH for complete neutralization.
In a second experiment, 2.000 grams of the acid is burned in excess oxygen,
producing 2.749 g of CO2 and 0.7503 grams of H2O.
Determine the molecular formula for the acid.
Answers to Unit 2 HW 3
1)
(a) net ionic: Zn + 2H+ → H2 + Zn+2
(b) net ionic: MgCO3 + 2H+ → CO2 + H2O + Mg+2
(c) net ionic: HC2H3O2 + HCO3- → H2O + CO2 + C2H3O2-
2)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
net ionic:
net ionic:
ne tionic:
net ionic:
net ionic:
net ionic:
net ionic:
3) MaVa = MbVb
4)
2H+ + Mg(OH)2 → 2H2O + Mg+2
3OH- + H3PO4 → 3H2O + PO4-3
HC2H3O2 + OH- → H2O + C2H3O2H2CO3 + 2OH- → 2H2O + CO3-2
2H+ + Ba(OH)2 → 2H2O + Ba+2
HF + OH- → H2O + FNO2- + H+ → HNO2
(0.8035 M) (10.00 mL) = (0.1105 M) (Vb)
Vb = 72.71 mL
mol NaOH = mol KHP = 1.004 g · 1 mol KHP = 0.004916 mol
204.23 g
[NaOH] = mol NaOH = 0.004916 mol = 0.131 M
L
0.0376 L
MaVa = MbVb
(Ma)(10.00 mL) = (0.131 M)(22.6 mL)
Ma = 0.295 M
5) mol acid = mol NaOH = (0.0250 L)(0.500 mol/L) = 0.0125 mol
Molar mass of acid = 2.20 g/0.0125 mol = 176 g/mol
Molar mass/empirical mass = 176 g/mol = 2
88.07 g/mol
Molecular formula = C6H8O6
6) Empirical formula = C22H20O11
Mol acid = mol NaOH = (0.01802 L)(0.0406 mol/L) = 7.32 X 10-4 mol
Molar mass of acid = 0.3602 g/ 7.32 X 10-4 mol = 492 g/mol
Molecular formula = empirical formula = C22H20O11
7) 3NaOH + H3X → Na3X + 3H2O
Mol acid = 1/3 mol NaOH = 1/3 (0.0372 L) (0.105 mol/L) = 0.00130 mol
Molar mass of acid = 0.250 g/ 0.00130 mol = 192 g/mol
Molecular Formula = Empirical formula: C6H8O7