CHEM 210 Exam 1 Review Questions and Correct Answers, With Complete Solution 2024.
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CHEM 210
CHEM 210 Exam 1 Review Questions and Correct Answers, With Complete Solution 2024.
What is a BL acid? BL base?
Acid is a proton donor while a base is proton acceptor
List the 6 strong acids
HClO4, HCl, HBr, HI, HNO3, H2SO4
List the 6 strong bases
LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
...
chem 210 exam 1 review questions and correct answe
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CHEM 210 Exam 1 Review Questions and
Correct Answers, With Complete Solution 2024.
What is a BL acid? BL base?
Acid is a proton donor while a base is proton acceptor
List the 6 strong acids
HClO4, HCl, HBr, HI, HNO3, H2SO4
List the 6 strong bases
LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
What is the eq. for the dissociation of a weak acid?
HA <=> H+ + A-
Ka = ([H+][A-])/[HA]
What is the eq. for base hydrolysis?
B + H2O <=> BH+ + OH-
Kb = ([BH+][OH-])/[B]
What is the equation for pH and pOH?
pH = -log[H+]
pH + pOH = -log(Kw) = 14
pOH = -log10 [OH-]
How would you convert pH to pOH and vice versa?
pH = 14 - pOH
pOH = 14 - pH
How would you convert pH to [H3O+]?
pH = - log[H3O+]
[H3O+] = 10^-pH
How would you convert [H3O+] to [OH-]?
Kw = 1 x 10^-14 = [H3O+][OH-]
Calculate the [OH-] in 6 M HNO3 and [H3O+] in 6.0 M NaOH at 25 C.
Kw = [OH][H3O] = 1 x 10^-14
(1 x 10^-14) / 6.0 M = 1.7 x 10^-15
The pH of a seawater sample is 8.30 what is [H3O+]?
[H3O+] = 10^-pH
10^-8.3 = 5.01 x 10^-9
An acid at a pH of 0 has an OH concentration of?
,1 x10^-14
and an H+ concentration of 1
What is the Ka and pKa range for a very strong acid?
Weak acid?
Extremely weak?
very strong: ka >0.1 pka <1
weak acid 10^-6 - 10^-3 pka 3-5
extremely < 10^-15 pka>15
Salt is the product of what rxn?
Acid-base
usually strong electrolytes that completely dissociate in water
Strong acid - strong acid
strong acid - weak base
strong base - strong base
weak acid - weak base
Calculate the pH of 0.010 M solution of sodium hypochlorite (NaClO) The Ka is 2.9
x 10^-18
Kw = Kb x Ka
Kb = Kw/ka
kb = 3.45 x 10^-7
R ClO- + H2O <=> OH- + HClO
I 0.010
C-X+X+X
E 0.010 - X X X
0 = X^2 - 3.45 X 10^-7 X + 3.45 X 10^-9
quadratic eq.
x = 5.86 x 10^-5 = [OH]
-log[OH] = pOH
pH = 14 - pOH
pH = 9.77
What eq. are useful when working with weak acids and bases?
,HA <=> H+ + A-
ka = [H][A]/[HA]
B + H2O <=> BH + OH
Kb = [BH][OH]/[B]
Calculate [H3O], pH, [HA], [A-], [OH], and alpha of .100 M propanoic acid
(CH3CH2CO2H)
pKa = 4.87 Ka = 1.34 x 10^-5
HA (aq) + H2O <-> H3O + A-
F-xxx
Ka = [H3O][A]/[HA] =
x^2 /(F-x)
[H3O] = [A-] = x
1.34 x 10^-5 = x^2 /(0.010-x)
quadratic eq.
x = 1.151 x 10^-3 = [H3O]
pH = -log(H3O) = 2.94
pOH = 14 - 2.94
[A-] = [H3O]
[HA] = F - x = 0.01 - 1.151 x 10^-3
alapha = x/ F (100) = 1.151 x 10^-3/.01 (100)
Calculate the [OH], [H3O], pH [BH], [B], and alpha of 0.1 M CH3NH2 Ka = 2.31 x 10
-11
B (aq) + H2O <-> BH + OH-
F-xxx
Kb = [BH][OH]/[B] =
x^2 /(F-x)
[BH] = [OH] = x
, kb = kw/ka = 4.33 x 10^-4
x = (-Kb +sq rt(kb)^2 + 4(Kb)(F))/2
x = 6.367 x 10^-3 = [OH] = [BH]
[H3O] = Kw/[OH]
pH = -log [H3O]
alpha = 6.367 x 10^-.1 (100)
[B] = 0.1 - 6.367 x 10^-3
What is charge balance?
The sum of the positive charges in solution equals the sum of the negative charges in
solution
n1[C1] + n2[C2] = m1[A1] + m2[A2]
C: concentration of cation
n = charge of cation
A: concentration of anion
m = charge of anion
What is mass balance?
The quantity of all species in a solution containing a particular atom must equal the
amount of that atom delivered to the solution
What are the steps to the systematic approach to solving eq. problems?
1) Write all relevant eq. rxns and eq. constant expressions
2) count the unique species appearing in the eq. constant expressions; these are the
unknowns. If the number of unknowns equalts the number of eq. constant expressions
then have enough information to solve. If not add a mass balance eq. and or charge
balance eq. Continue adding until the number of equations = the number of unknowns
3) combine eq. and solve for one unknown
4) check assumptions
What are helpful equations when dealing with weak acid eq. ?
H2O <=> H+ + OH-
[H+] = [A-][OH-]
Kw =[H+][OH-]
F = [A=] + [HA]
[H] ~= [A-]
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