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Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer based on the definition given in the module:
Your Answer: A. acid B. base C. salt D. salts
Question 1 Not^ yet^ graded^ /^0 pts
CHEM 104 MODULE 2- PROBLEM SET- GENERAL CHEMISTRY 2 WITH LAB-
SCHIREN PORTAGE LEARNING 2023 - 2024
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Submitted Jun 21 at 6:01pm This attempt took 1,112 minutes.
(a) HC 6 H 5 O 2 (b) Be(OH) 2 (c) Mn(ClO 2 ) 2 (d) AgCl (e) Sn(OH) 4 (f) H 2 CrO 4 (g) PbO 2 (h) Ni(OH) 4 (i) H 3 AsO 4 (j) HC 2 H 3 O 2
E. base
F.acid
G. salt
H. base
I. acid
J. acids
Identify the following unknown substances as ACID, BASE or SALT on the basis of their property.
Question 2 Not^ yet^ graded^ /^0 pts
(a) Solution of A has a pH of 4.
(b) Solution of B causes litmus to turn blue (c) Substance C has a bitter taste
(d) Solution of D has a pH of 8.
(e) Substance E has a sour taste (f) Solution of F causes litmus to turn red
Your Answer:
A. acid
B. base
C. base
D. base
E. acid
F. acid
Identify the following unknown substances as ACID, BASE or SALTon the basis of their property.
(a) A is an acid since its pH is < 7 (4.6) (b) B is a base. Bases causes litmus to turn blue
(c) C is a base. Bases have a bitter taste
(d) D is a base since its pH is > 7 (8.9) (e) E is an acid. Acids have a sour taste
(f) F is an acid. Acids causes litmus to turn red
In the following reaction identify each reactant as a Brønsted-Lowry acid or a Brønsted- Lowry base and explain your answers.
HCl + NH 3 → NH 4 + + Cl-
Your Answer:
HCL - acid - donates H+
NH3 - base - accepts H+
Question 4 Not^ yet^ graded^ /^0 pts
In the following reaction identify each reactant as an Arrhenius acid or an Arrhenius base and explain your answers.
NaOH + HCl → NaCl + H 2 O
Your Answer: NaOH = base because OH HCL = acid because H NaCl is a salt H20= base
HCl = Arrhenius acid because it provides a H+ NaOH = Arrhenius base because it provides a OH-
In the following reaction label each material as to whether it is stronger acid, stronger base, weaker acid or weaker base and explain your answer.
Question 6 Not^ yet^ graded^ /^0 pts
Question 5 Not yet graded / 0 pts
In the following reaction identify each reactant as a Lewis acid or a Lewisbase and explain your answers.
Na 2 O: + SO 3 → Na 2 SO 4
Your Answer:
Lewis acids are the species that can accept the pair of non bonding electrons whereas the lewis bases are the species that donate the pair ofnon bonding electrons. Above the pair of electrons accepted by SO3 , soSO3 is the lewis acid and Na2O: donates the electron pair so it is the lewis base.
Na 2 O = Lewis base because it is an electron-pair donor SO 3 =
Lewis acid because it is an electron-pair acceptor
HCl = Brønsted-Lowry acid because it is a H+^ donor NH 3 = Brønsted-Lowry base because it is a H+^ acceptor
Give the conjugate base for each substance listed below:
(a) H 2 SO 4 (b) HCO 3 - (c) (d)
HC 2 H 3 O 2
H 2 S
Your Answer:
NH
HCO3-
CN-
OH-
Question 8 Not^ yet^ graded^ /^0 pts
b. H2CO c. HCN d. H2O
A conjugate acid has one more H+: (a) NH 3 → NH 4 + (b)
(d)
HCO 3
CN-
OH-
(c) →
H 2 CO 3
HCN
→ H 2 O
Choose and explain which of the following pairs is the stronger acid.
(a) (b) (c) (d)
HNO 3 or HNO 2 H 2 S or H 2 Te H 3 PO 4 or H 2 SO 4 HCl or HBr
Your Answer:
a. HNO3 stronger acid. higher ka
b. H2Te is a stronger acid ka = 2 - 3x10^- 3
c. H2SO4 stronger acid cause ka value is higher
d. HBr stronger acid because same as above and bond length higher
Question 9 Not^ yet^ graded^ /^0 pts
A conjugate base has one less H+: (a) (b)
H 2 SO 4
HCO 3
HC 2 H 3 O 2
H 2 S
→ HSO 4 -
- → CO 3 - 2
(c) → - (d) →
C 2 H 3 O 2
HS-
Complete the following chemical reactions:
(a) HCN + NaOH →
(b) Zn + 2 HCl →
(c) (^) Al 2 O 3 + 3 H 2 SO 4 →
(d) 2 HBr + CaCO 3 →
(e) 2 NaOH + SO 3 →
Your Answer:
A. NaC + H2O
B. ZnCl2+H
Question 11 Not^ yet^ graded^ /^0 pts
D. MgCL2 + 2H2O E. 3Na+ +PO3-4 +3H2O
(a) (b)
K 2 O + H 2 O → 2 K+^ + 2 OH-
CaO + 2 HCl → Ca+2^ + 2 Cl-^ + H 2 O (c) (d) (e)
P 2 O 5 +
Mg(OH) 2 PO(OH) 3
3 H 2 O → 2 H 3 PO 4
- 2 HCl → Mg+2^ + 2 Cl-^ + 2 H 2 O
- 3 NaOH → 3 Na+^ + PO 4 -^2 + 3 H 2 O
Show the calculation of the [H+] of a solution whose [OH-] = 3.89 x 10 -^3 M. Explain whether this solution is acidic or basic.
Your Answer:
[OH+] [OH-] = 1.0x10^- 14
[H+] = 1.0x10^-14/[OH-]
Given
[H+] = 3.89x10^-14/ 3.89x10^- 3
=0.257x10^- 11
Question 12 Not^ yet^ graded^ /^0 pts
C. Al2(SO4)3 + 3H2O D. CaBr2+ CO2+H2O E. Na2SO4 + H2O
(a) HCN + NaOH → Na+^ + CN-^ + H 2 O
(b) Zn + 2 HCl → Zn+2^ + 2 Cl-^ + H 2 (g) ↑
(c) Al 2 O 3 + 3 H 2 SO 4 → 2 Al+3^ + 3 SO 4 -^2 + 3 H 2 O
(d) 2 HBr + CaCO 3 → CaBr 2 + (H 2 CO 3 ) → H 2 O + CO 2 (g) ↑
(e) 2 NaOH + SO 3 → Na 2 SO 4 + H 2 O
Show the calculation of the [H+] of a solution whose pH = 4.72.
Your Answer:
ph = 4.
= log [H+] = 3.
Question 15 Not^ yet^ graded^ /^0 pts
pH = 14 - pOH = 14 - 2.870 = 11.
[OH-] = [NaOH] = 0.00135 M pOH = - log (0.00135) =
Your Answer: [OH-] = 0. pOH = - log [OH-]
- log [OH]= - log (0. = pOH = 2. we know, ph + poh = 14 =ph + 2.86 = 14 = ph = 14 - 2.86 = 11. so pH 0.00135 is a strong base
Show the calculation of the pH of a 0.00135 M solution of the strong base NaOH.
Question 14 Not^ yet^ graded^ /^0 pts
Show the calculation of the [OH-] of a solution whose pH = 10.34.
Your Answer:
pH + pOH= 14
P OH = 14 - pH
= 14 - 10.34 = 3.
pOH = - log 10 [OH]
log 10 - [OH-] = - pOH
= 10 - 3.
= 0.0002187M
[OH-] = 2.19 x 10-4 M
Question 16 Not^ yet^ graded^ /^0 pts
= [H+] = 10 ^4.
= [H+] = 1.90 x10 ^- 3 log [H+] = 1.9 x10 ^- 5
pH = 4. 4.72) = 1.9 x 10 -^5
[H+] = antilog (- pH) [H+] = antilog (-
Question 18 Not yet graded / 0 pts
In the titration of 0.127 grams of Al(OH) 3 in the presence of bromothymolblue indicator, the blue solution turns yellow after 22.3 mL of H 2 SO 4 of unknown concentration is added. Show the calculation of the molarity ofthe H 2 SO 4 solution?
Your Answer:
Now the reaction between base Al(OH)3 and acid H2SO4 is-
3H2SO4 + 2Al(OH)3 -------- > Al2(SO4)3 + 6H2O
Thus 2 moles of Al(OH) 3 is required to neutralize 3 moles of 3H2SO4.
0.127 grams used of Al(OH) 3. And number of moles of Al(OH)3 presentis-
weight of Al(OH)3 taken / molar mass of Al(OH)
= 0.127 grams / 78 g/mol
= 0.0016 mols
0.0016 mols of Al(OH) 3 can neutralize = (3/2) * 0.0016 mols of H2SO4.
= 0.0024 mols
H2SO4 present in 22.3 mL of solution.
concentration of H2SO4is = (no. of moles / Volume ) * 1000
= (0.0024 moles / 22.3 mL ) * 1000
( H Cl is the acid and Na OH is the base) Ma x mLa / 1000 x Sa / Sb = Mb x mLb / 1000 Ma x 26.3 / 1000 x 1 / 1 = 0.132 x 28.4 / 1000 MHCl = 0.143 M
= 0.107 M
