Access over 20 million homework & study documents

Experiment 5

Content type
User Generated
Subject
Chemistry
Type
Homework
Rating
Showing Page:
1/3
Experiment 5 - Pre Lab
1. Define the following terms:
a) Bronsted acid and base - acid and bases are defined by the transfer of a proton
(H+), where the acid is a proton donor and the base is a proton acceptor.
b) Arrhenius acid and base - an acid is any compound dissolved in water that has a
hydrogen ion concentration greater than water, while a base has accepts
hydrogen ions.
2. Write the net ionic equation for the following ions which undergo hydrolysis
a) NO2-
NO2- (aq) + H2O (l) HNO2 (aq) + OH- (aq)
b) ClO3-
ClO3- (aq) + H2O (l) HClO3 (aq) + OH- (aq)
c) Zn 2+
Zn 2+ (aq) + H2O (l) ZnOH+ + H+
3. Identify the following salts as acidic, basic or neutral:
Use the table found in the lab manual, to determine.
a) NaClO3 - Na+ is neutral and ClO3- is neutral, so the salt is neutral
b) KCN - K+ is neutral and CN- is basic, so the salt is basic
c) Na2SO3 - Na+ is neutral and SO3 2- is basic, so the salt is basic.
4. Aniline is a weak organic base with the formula C6H5NH2. The anilinium ion has
the formula C6H5NH3+ and its Ka is 2.3 x 10^-5.
a) Write the chemical reaction showing the hydrolysis of the anilinium ion.
C6H5NH3 +(aq) + H2O (l) C6H5NH2 (aq) + H3O+ (aq)
b) Calculate the pH of a 0.0400 M of anilinium chloride
Set up an ICE table
C6H5NH3 +(aq) + H2O (l) C6H5NH2 (aq) + H3O+ (aq)
I
0.0400 M
XXXX
0 M
0 M
C
-x
XXXX
+x
+x

Sign up to view the full document!

lock_open Sign Up
Showing Page:
2/3
E
0.04 -x
XXXX
x
x
Ka = x^2 / (0.04-x) = 2.3e-5
x^2 = 9.2e-7 - 2.3e-5 x
x^2 + 2.3e-5 x - 9.2e-7 = 0
x = 0.0009477352422633355
pH = -log (0.0009477352422633355) = 3.02
5. Find the pH of a 0.20 M solution of sodium propionate, NaC3H5O2. Ka of propionic
acid, HC3H5O2, is 1.34 x 10^-5 (Hint: Kw = Ka*Kb)
C3H5O2- (aq) + H2O HC3H5O2 + OH -
I
0.2 M
XXXX
0 M
C
-x
XXXX
+x
E
0.2 -x
XXXX
x
Find Kb for the equation above
Kb = Kw / Ka = 1e-14 / 1.34e-5
Kb = 7.462686e-10
Kb = x^2 / (0.2-x) = 7.462686e-10
x^2 = 1.4925e-10 - 7.46e-10x
x^2 +7.46e-10 x - 1.4925e-10 = 0
x = 0.00001221657002424949 = [OH-]
pOH = - log (0.00001221657002424949)
pOH = 4.91

Sign up to view the full document!

lock_open Sign Up
Showing Page:
3/3

Sign up to view the full document!

lock_open Sign Up
Unformatted Attachment Preview
Experiment 5 - Pre Lab 1. Define the following terms: a) Bronsted acid and base - acid and bases are defined by the transfer of a proton (H+), where the acid is a proton donor and the base is a proton acceptor. b) Arrhenius acid and base - an acid is any compound dissolved in water that has a hydrogen ion concentration greater than water, while a base has accepts hydrogen ions. 2. Write the net ionic equation for the following ions which undergo hydrolysis a) NO2NO2- (aq) + H2O (l) ↔ HNO2 (aq) + OH- (aq) b) ClO3ClO3- (aq) + H2O (l) ↔ HClO3 (aq) + OH- (aq) c) Zn 2+ Zn 2+ (aq) + H2O (l) ↔ ZnOH+ + H+ 3. Identify the following salts as acidic, basic or neutral: Use the table found in the lab manual, to determine. a) NaClO3 - Na+ is neutral and ClO3- is neutral, so the salt is neutral b) KCN - K+ is neutral and CN- is basic, so the salt is basic c) Na2SO3 - Na+ is neutral and SO3 2- is ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Anonymous
Goes above and beyond expectations!

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4