Weak Acids & Weak Acids & Weak BasesWeak Bases
ReviewReview Try the next two questions to see what Try the next two questions to see what
you remember you remember
The pH of rainwater collected on a particular day was 4.82. What is the H+ ion concentration of the rainwater?
pH = -log [H+]
[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M
The OH- ion concentration of a blood sample is 2.5 x 10-7
M. What is the pH of the blood?
pH + pOH = 14.00
pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60
pH = 14.00 – pOH = 14.00 – 6.60 = 7.40
New TopicNew Topic Weak Acids & Weak BasesWeak Acids & Weak Bases
Strong Electrolyte – 100% dissociation
NaCl (s) Na+ (aq) + Cl- (aq)
Weak Electrolyte – not completely dissociated
CH3COOH CH3COO- (aq) + H+ (aq)
Strong Acids are strong electrolytes
HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)
H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)
HF (aq) + H2O (l) H3O+ (aq) + F- (aq)
Weak Acids are weak electrolytes
HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)
HSO4- (aq) + H2O (l) H3O+ (aq) + SO4
2- (aq)
H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)
Strong Bases are strong electrolytes
NaOH (s) Na+ (aq) + OH- (aq)
KOH (s) K+ (aq) + OH- (aq)
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
F- (aq) + H2O (l) OH- (aq) + HF (aq)
Weak Bases are weak electrolytes
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Conjugate acid-base pairs:• The conjugate base of a strong acid has no measurable
strength.
• H3O+ is the strongest acid that can exist in aqueous solution.
• The OH- ion is the strongest base that can exist in aqueous solution.
Strong Acid Weak Acid
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
Start
End
0.002 M
0.002 M 0.002 M0.0 M
0.0 M 0.0 M
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)
Start
End
0.018 M
0.018 M 0.036 M0.0 M
0.0 M 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Weak Acids (HA) and Acid Ionization Constants
HA (aq) H+ (aq) + A- (aq)
Ka =[H+][A-][HA]
Ka is the acid ionization constant
Ka
weak acidstrength
Important PointsImportant Points When weak acids and bases are added to When weak acids and bases are added to
water the concentration of [Hwater the concentration of [H33OO++] and ] and [OH[OH--] is not the same as the concentration ] is not the same as the concentration of the dissolved acid. of the dissolved acid.
The concentration can be determined The concentration can be determined using the using the equilibrium constantequilibrium constant values values for acids and bases termed Kfor acids and bases termed Kaa and K and Kbb. .
Important PointsImportant Points KKaa values are the equilibrium constant values are the equilibrium constant
expressions for the dissolving of various acids expressions for the dissolving of various acids in water. in water.
The larger the value of KThe larger the value of Kaa the stronger the the stronger the acids are. acids are.
The same is true of bases and KThe same is true of bases and Kbb. The . The following link contains following link contains a table of acidsa table of acids and their and their KKaa values values
Calculating [HCalculating [H33OO++] and [OH] and [OH--] of Weak ] of Weak
Acids & BasesAcids & Bases We can calculate the concentration of the We can calculate the concentration of the
[H[H33OO++] or [OH] or [OH--] for a weak acid or base using K] for a weak acid or base using Kaa and Kand Kbb values and the following procedure. values and the following procedure.
The procedure assumes that the concentration The procedure assumes that the concentration of the dissolved acid or base does not change of the dissolved acid or base does not change significantly when the weak acid dissolves.significantly when the weak acid dissolves.
Procedure for Calculating [HProcedure for Calculating [H33OO++] and ] and [OH[OH--] of Weak Acids & Bases] of Weak Acids & Bases
1.1. Write the dissolving reaction for the weak acid. Write the dissolving reaction for the weak acid. 2.2. These are typically These are typically monoproticmonoprotic (acids or (acids or
bases) that produce only one Hbases) that produce only one H33OO++ ion or OH ion or OH-- ion. ion.
3.3. Write the KWrite the Kaa or K or Kbb expression for the acid or expression for the acid or base. base.
4.4. Calculate the concentration of the acid or base Calculate the concentration of the acid or base in moles per liter.in moles per liter.
5.5. Using a table of acids and bases locate the Using a table of acids and bases locate the value of Kvalue of Kaa or K or Kbb for the acid and base and for the acid and base and then use the following formulas to calculate the then use the following formulas to calculate the [H[H33OO++] or [OH] or [OH--]]
[H[H33OO++]]22 = [Acid]* K = [Acid]* Kaa
[OH[OH--]]2 2 = [Base] *K= [Base] *Kbb
6.6. Using the KUsing the Kww formula calculate the other formula calculate the other concentration or [OHconcentration or [OH--] or [H] or [H33OO++]]
Problem 1:Problem 1: Calculate the concentration of [HCalculate the concentration of [H33OO++] and [OH] and [OH--] ]
ions in a 0.0056 mol/l solution of HCN . ions in a 0.0056 mol/l solution of HCN . (K(Kaa value = 6.2 X10 value = 6.2 X10-10-10) )
Strategy:Strategy:1.1. Write the dissolving reaction: Write the dissolving reaction:
HCNHCN(aq) (aq) + H+ H22O O (l)(l) ↔ H ↔ H33OO++(aq)(aq) + CN + CN-- (aq)(aq)
2.2. Write the equilibrium expression KWrite the equilibrium expression Kaa : : KKaa = [H = [H33OO++]*[CN]*[CN--] / [HCN]] / [HCN]
3.3. Determine the concentration of the weak acid , Determine the concentration of the weak acid , [HCN] = 0.0056 mol/l [HCN] = 0.0056 mol/l
3.3. Using Using a table of acids a table of acids determine that the Kdetermine that the Ka a
value is = 6.2 X10value is = 6.2 X10-10-10
4.4. Calculate the [HCalculate the [H33OO++] by substituting information ] by substituting information into the formula to get into the formula to get [H[H33OO++]]22 = (0.0056 mol/L)(6.2 X10 = (0.0056 mol/L)(6.2 X10-10-10))[H[H33OO++] = 1.86 X10] = 1.86 X10-6-6 mol/L mol/L
5.5. Using the KUsing the Kww formula calculate the [OH formula calculate the [OH--] ; ] ; remember Kremember Kww = 1.0 X10 = 1.0 X10-14-14
[OH[OH--] = K] = Kww / [H / [H33OO++]][OH[OH--] = 1.0 X10] = 1.0 X10-14-14/ 1.86 X 10/ 1.86 X 10-6-6 mol/L mol/L[OH[OH--] = 5.37 X10] = 5.37 X10-9-9 mol/L mol/L
Important Note:Important Note: The procedure and formula used above only The procedure and formula used above only
work if the concentration of the weak acids and work if the concentration of the weak acids and bases is not significantly changed when it bases is not significantly changed when it dissolves. dissolves.
This is true when the [Acid] ÷ KThis is true when the [Acid] ÷ Kaa > 500 > 500
Work through the following questions by Work through the following questions by creating ICE tables that support this same creating ICE tables that support this same logic or strategy!logic or strategy!
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]
= 7.1 x 10-4
HF (aq) H+ (aq) + F- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50 0.00
-x +x
0.50 - x
0.00
+x
x x
Ka =x2
0.50 - x= 7.1 x 10-4
Ka x2
0.50= 7.1 x 10-4
0.50 – x 0.50The ratio is > 500,
x2 = 3.55 x 10-4 x = 0.019 M
[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72
[HF] = 0.50 – x = 0.48 M
When can I use the approximation?
0.50 – x 0.50Ka << 1
When the ratio of initial concentration/Ka is greater than 500.
x = 0.5 0.50 M7.1 x 10-4
= 714 which is > 500
What is the pH of a 0.05 M HF solution (at 250C)?
0.05
7.1 X 10-4 = 71.4 which is < 500
We can not neglect the “x” value. We must solve for x exactly using quadratic equation or method of successive approximation.
Solving weak acid ionization problems:
1. Identify the major species that can affect the pH.
• In most cases, you can ignore the auto-ionization of water.
• Ignore [OH-] because it is determined by [H+].
2. Use ICE to express the equilibrium concentrations in terms of single unknown x.
3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.
4. Calculate concentrations of all species and/or pH of the solution.
What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122 0.00
-x +x
0.122 - x
0.00
+x
x x
Ka =x2
0.122 - x= 5.7 x 10-4
Ka x2
0.122= 5.7 x 10-4
0.122 – x 0.122Ka << 1
0.1225.7 X 10-4
= 214.04This value is < 500
Approximation not valid.
Ka =x2
0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0
ax2 + bx + c =0-b ± b2 – 4ac
2ax =
x = 0.0081 x = - 0.0081
HA (aq) H+ (aq) + A- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122 0.00
-x +x
0.122 - x
0.00
+x
x x
[H+] = x = 0.0081 M pH = -log[H+] = 2.09
percent ionization = Ionized acid concentration at equilibrium
Initial concentration of acidx 100%
For a monoprotic acid HA
Percent ionization = [H+]
[HA]0
x 100% [HA]0 = initial concentration
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Weak Bases and Base Ionization Constants
Kb =[NH4
+][OH-][NH3]
Kb is the base ionization constant
Kb
weak basestrength
Solve weak base problems like weak acids except solve for [OH-] instead of [H+].
Chemistry In Action: Antacids and the Stomach pH Balance
NaHCO3 (aq) + HCl (aq)
NaCl (aq) + H2O (l) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq)
MgCl2 (aq) + 2H2O (l)
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