CHEM134 COOL ACTUAL SPRING 2024 DIGITAL ACID BASE CHEMISTRY - FULLY SOLVE100%

CHEM134 COOL ACTUAL SPRING 2024 DIGITAL ACID BASE CHEMISTRY - FULLY
SOLVE100%

Question 1
Q: Define an acid and a base according to the Brønsted-Lowry theory.

A: According to the Brønsted-Lowry theory:

● An acid is a proton (H⁺) donor.
● A base is a proton (H⁺) acceptor.

Question 2
Q: Write the balanced equation for the neutralization reaction between hydrochloric acid (HCl)
and sodium hydroxide (NaOH).

A: The balanced equation for the neutralization reaction is: HCl(aq)
+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl} (aq) + \text{NaOH} (aq) \rightarrow \
text{NaCl} (aq) + \text{H}_2\text{O} (l)HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Question 3
Q: Calculate the pH of a 0.01 M HCl solution.

A: HCl is a strong acid and completely dissociates in water: HCl→H++Cl−\
text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-HCl→H++Cl−

The concentration of H+\text{H}^+H+ ions is 0.01 M. pH is calculated using the
formula: pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]pH=−log[H+] pH=−log⁡(0.01)\
text{pH} = -\log(0.01)pH=−log(0.01) pH=2\text{pH} = 2pH=2

Question 4
Q: What is the pH of a 0.1 M solution of acetic acid (CH₃COOH), given that the
KaK_aKa of acetic acid is 1.8×10−51.8 \times 10^{-5}1.8×10−5?

A: For acetic acid, we set up the following equilibrium expression: CH3COOH⇌H+
+CH3COO−\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\
text{COO}^-CH3COOH⇌H++CH3COO−

The equilibrium expression is: Ka=[H+][CH3COO−][CH3COOH]K_a = \frac{[\
text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}Ka=[CH3
COOH][H+][CH3COO−]

Assuming xxx is the concentration of H+\text{H}^+H+ ions at equilibrium, we
have: Ka=x20.1−xK_a = \frac{x^2}{0.1 - x}Ka=0.1−xx2

,Since KaK_aKa is very small, we can approximate 0.1−x≈0.10.1 - x \approx
0.10.1−x≈0.1: 1.8×10−5=x20.11.8 \times 10^{-5} = \frac{x^2}
{0.1}1.8×10−5=0.1x2 x2=1.8×10−6x^2 = 1.8 \times 10^{-6}x2=1.8×10−6
x=1.8×10−6x = \sqrt{1.8 \times 10^{-6}}x=1.8×10−6 x≈1.34×10−3x \approx
1.34 \times 10^{-3}x≈1.34×10−3

Thus, the H+\text{H}^+H+ concentration is 1.34×10−3M1.34 \times 10^{-3}
M1.34×10−3M. The pH is: pH=−log⁡(1.34×10−3)\text{pH} = -\log(1.34 \times
10^{-3})pH=−log(1.34×10−3) pH≈2.87\text{pH} \approx 2.87pH≈2.87

Question 5
Q: What is the difference between a strong acid and a weak acid?

A:

● A strong acid completely dissociates in water, meaning it releases all its hydrogen ions
(H⁺) into the solution. Examples include HCl, HNO₃, and H₂SO₄.
● A weak acid only partially dissociates in water, meaning only a fraction of its hydrogen
ions are released into the solution. Examples include acetic acid (CH₃COOH) and formic
acid (HCOOH).

Question 6
Q: Determine the pH of a 0.05 M NaOH solution.

A: NaOH is a strong base and completely dissociates in water: NaOH→Na++OH−\
text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-NaOH→Na++OH−

The concentration of OH−\text{OH}^-OH− ions is 0.05 M. To find the pH, we first
find the pOH: pOH=−log⁡[OH−]\text{pOH} = -\log[\text{OH}^-]pOH=−log[OH−]
pOH=−log⁡(0.05)\text{pOH} = -\log(0.05)pOH=−log(0.05) pOH=1.3\text{pOH} =
1.3pOH=1.3

Using the relationship between pH and pOH: pH+pOH=14\text{pH} + \text{pOH}
= 14pH+pOH=14 pH=14−1.3\text{pH} = 14 - 1.3pH=14−1.3 pH=12.7\text{pH} =
12.7pH=12.7

Feel free to ask if you need more questions or if there's anything specific you'd like to cover!

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Question 7
Q: What is the conjugate base of HSO₄⁻?

, A: The conjugate base of HSO₄⁻ is SO₄²⁻.

Question 8
Q: What is the pH of a 0.1 M solution of HNO₃?

A: HNO₃ is a strong acid and completely dissociates in water: HNO3→H++NO3−\
text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^-HNO3→H++NO3−

The concentration of H+\text{H}^+H+ ions is 0.1 M. pH is calculated using the
formula: pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+]pH=−log[H+] pH=−log⁡(0.1)\
text{pH} = -\log(0.1)pH=−log(0.1) pH=1\text{pH} = 1pH=1

Question 9
Q: Write the balanced equation for the dissociation of sulfuric acid (H₂SO₄) in water.

A: The balanced equation for the dissociation is: H2SO4(aq)→2H+(aq)+SO42−(aq)\
text{H}_2\text{SO}_4 (aq) \rightarrow 2\text{H}^+ (aq) + \text{SO}_4^{2-}
(aq)H2SO4(aq)→2H+(aq)+SO42−(aq)

Question 10
Q: What is the K_b of NH₃ if the KaK_aKa of NH₄⁺ is 5.6×10−105.6 \times 10^{-
10}5.6×10−10?

A: The relationship between KaK_aKa and KbK_bKb for a conjugate acid-base pair is:
Ka×Kb=KwK_a \times K_b = K_wKa×Kb=Kw where KwK_wKw is the ion-product
constant for water (1.0×10−141.0 \times 10^{-14}1.0×10−14 at 25°C).

Kb=KwKaK_b = \frac{K_w}{K_a}Kb=KaKw Kb=1.0×10−145.6×10−10K_b = \
frac{1.0 \times 10^{-14}}{5.6 \times 10^{-10}}Kb=5.6×10−101.0×10−14
Kb=1.79×10−5K_b = 1.79 \times 10^{-5}Kb=1.79×10−5

Question 11
Q: Calculate the pH of a 0.02 M solution of Ba(OH)₂.

A: Ba(OH)₂ is a strong base and completely dissociates in water: Ba(OH)2→Ba2+
+2OH−\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-Ba(OH)2
→Ba2++2OH−

The concentration of OH−\text{OH}^-OH− ions is 2×0.02=0.042 \times 0.02 =
0.042×0.02=0.04 M. To find the pOH: pOH=−log⁡[OH−]\text{pOH} = -\log[\
text{OH}^-]pOH=−log[OH−] pOH=−log⁡(0.04)\text{pOH} = -\
log(0.04)pOH=−log(0.04) pOH=1.4\text{pOH} = 1.4pOH=1.4

Using the relationship between pH and pOH: pH+pOH=14\text{pH} + \text{pOH}
= 14pH+pOH=14 pH=14−1.4\text{pH} = 14 - 1.4pH=14−1.4 pH=12.6\text{pH} =
12.6pH=12.6