Topic 2: Questions

Question 1

Write a fully balanced chemical equation for the reactions between hydrogen bromide, HBr with water. Identify the acid, base, conjugate acid and conjugate base.

Answer 1

HBr + H2O → H3O+ + Br-
HBr: acid (donates proton H+)
H2O: base (accepts protons H+)
H3O+: conjugate acid (result of base accepting proton from acid)
Br-: conjugate base (result of acid donating proton to base)

Question 2

Write a fully balanced chemical equation for the reactions between hydrogen sulphide, H2S with water. Identify the acid, base, conjugate acid and conjugate base.

Answer 2

H2S + H2O → H3O+ + HS-
H2S: acid (donates proton H+)
H2O: base (accepts protons H+)
H3O+: conjugate acid (result of base accepting proton from acid)
HS-: conjugate base (result of acid donating proton to base)

Question 3

Write a fully balanced chemical equation for the reactions between hydrogen sulfate ion, HSO4- with water. Identify the acid, base, conjugate acid and conjugate base.

Answer 3

HSO4- + H2O → H3O+ + SO4^-2
HSO4-: acid (donates proton H+)
H2O: base (accepts protons H+)
H3O+: conjugate acid (result of base accepting proton from acid)
SO4^2-: conjugate base (result of acid donating proton to base)

Question 4

Write a fully balanced equation for the acid-base reaction between nitric acid, HNO3 and potassium hydroxide, KOH. Identify the acid, base, conjugate acid and conjugate base.

Answer 4

HNO3 + KOH → H2O + KNO3
Single-displacement reaction
HNO3: acid (donates proton H+)
KOH: base (accepts protons H+)
H2O: conjugate acid (result of base accepting proton from acid)
KNO3: conjugate base (result of acid donating proton to base)

Question 5

Write a fully balanced equation for the acid-base reaction between perchloric acid, HClO4 and ammonia, NH3. Identify the acid, base, conjugate acid and conjugate base.

Answer 5

HClO4 + NH3 → CLO4- + NH4+
Single-displacement reaction
HClO4: acid (donates proton H+)
NH3: base (accepts protons H+)
NH4+: conjugate acid (result of base accepting proton from acid)
CLO4-: conjugate base (result of acid donating proton to base)

Question 6

Identify the acid, base, conjugate acid and conjugate base of:
OH- + H2CO3 ⇌ H20 + HCO3-

Answer 6

OH- + H2CO3 ⇌ H20 + HCO3-
H2CO3: acid (donates proton H+)
OH-: base (accepts protons H+)
H2O: conjugate acid (result of base accepting proton from acid)
HCO3-: conjugate base (result of acid donating proton to base)

Question 7

Identify the acid, base, conjugate acid and conjugate base of:
HI + PO4^-3 → I- + HPO4^-2

Answer 7

H2CO3: acid (donates proton H+)
OH-: base (accepts protons H+)
H2O: conjugate acid (result of base accepting proton from acid)
HCO3-: conjugate base (result of acid donating proton to base)

Question 8

Write an equation for the chemical reaction between iodic acid, HIO3 and water, H2O.

Answer 8

HIO3 + H2O ⇌ IO3- + H3O+

Question 9

Classify chromic acid, H2CrO4 as monoprotic, diprotic, triprotic and write an equation to show the ionisation of H2CrO4 with water, H2O.

Answer 9

Diprotic: two protons are available to be donated
-protic (amount of protons/hydrogen that can be donated)
H2CrO4 + H2O → HCrO4- + H3O+
HCrO4- + H2O → CrO4^-2 + H3O+
To ionise diprotic acids there are two steps. For each step one proton is transferred to different water molecules.

Question 10

Balance the equation H2 + S8 → H2S then write to molar ratio for:
H2 : H2S
H2S : S8
H2 : S8

Answer 10

8H2 + S8 → 8H2S
H2 : H2S ratio is 8:8 –> 1:1
H2S : S8 ratio is 8:1
H2 : S8 ratio is 8:1

Question 11

1. Balance the equation C3H8 + O2 → H2O + CO2.
2. Calculate the amount of moles of water that can be produced with 28g of C3H8.
3. Calculate the amount of moles of water that can be produced with 25g of O2.
4. Identify the limiting reactant and explain why
   (complete this question in 5-8minutes)

Answer 11

1. Balanced equation: C3H8 + 5O2 → 4H2O + 3CO2
2. n(C3H8) = m / mr
   n(C3H8) = 28 / (12.01 x 3 + 1.008 x 8)
   n(C3H8) = 0.635
   n(H2O) = unknown(H2O) / known(C3H8) x mol of known (C3H8)
   n(H2O) = 4/1 x 0.635
   n(H2O) = 2.54mol of water can be produced with 28g if C3H8
3. n(O2) = m / mr
   n(O2) = 25 / (16 x 2)
   n(O2) = 3.125mol
   n(H2O) = unknown(H2O) / known(O2) x mol of known (O2)
   n(H2O) = 4/5 x 3.125
   n(H2O) = 2.5mol of water can be produced with 25g if O2
4. The limiting reactant is O2 because it produces less mols than C3H8 and thus will run out first

Question 12

Completely balance the REDOX equation

Cr2O7^-2 + Fe^+2 → Cr^+3 + Fe^+3

Answer 12

ox: 6(Fe^+2 → Fe^+3 + e-)
red: 6e- + 14H+ + Cr2O7^-2 → 2Cr^+3 + 7H2O
6e- + 14H+ + Cr2O7^-2 + 6Fe^+2 → 2Cr^+3 + 7H2O + 6Fe^+3 + 6e-
Completely balanced REDOX equation:
14H+ + Cr2O7^-2 + 6Fe^+2 → 2Cr^+3 + 7H2O + 6Fe^+3

Question 13

Calculate the pH of river water with [H+] = 0.1mmol L-1

Answer 13

pH = -log[H+]
m(milli) = 10^-3
[H+] = 0.1 x 10^-3
pH = -log[10^-3]

Question 14

Blood has a pH of 7.4. Calculate the hydrogen ion concentration of blood.

Answer 14

[H+] = 10^-pH
[H+] = 10^-7.4
[H+] = 3.9E-9
[H+] = 4 x 10^-9

Question 15

Identify the oxidation numbers and half-reaction for

2Al + 3H202 → Al2O3 + 3H2O

Answer 15

2Al + 3H2O2 → Al2O3 + 3H2O
(0) (-1) (+3) (-2)

oxidation: 2Al → Al2O3
reduction: 3H2O2 → 3H2O