Exam (elaborations)
CHEM134 COOL ACTUAL SPRING 2024 DIGITAL ACID BASE CHEMISTRY - FULLY SOLVE100%
- Course
- Institution
CHEM134 COOL ACTUAL SPRING 2024 DIGITAL ACID BASE CHEMISTRY - FULLY SOLVE100%
[Show more]
Content preview
CHEM134 COOL ACTUAL SPRING 2024 DIGITAL ACID BASE CHEMISTRY - FULLYSOLVE100%
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!
4o
50 MORE
ChatGPT
Sure, here are 50 more questions and answers on acid-base chemistry:
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
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller smartguide. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $4.89. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
75632 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now