Find the Ph of a 0.120 M Hf Solution

Titrations

We will look at four types of titrations:

�� I.� A strong acid with a strong base

�� II.� A weak acid with a strong base

�� III.� A strong acid with a weak base

�� IV.� Titration of a polyprotic acid

I.� A strong acid titrated with a strong base

We will calculate the pH of 25 mL of 0.1 M HCl titrated with 0.1 M NaOH.� At each point in the titration curve, we will need to determine two quantities, the concentration of H⁺ remaining in the solution, and the volume of the solution.� From these, we can calculate the [H⁺] and, from that, the pH.�

1.� �� The initial pH, before the addition of strong base.�

�� In dealing with strong acids and bases, assume that they dissociate completely in water.

�� [H⁺] = [HCl]

�� In this example, we have a 0.1 M solution of HCl, so the [H⁺] = 0.1 M

�� - log (0.1) = 1 = pH

2.�� The pH after addition of 10 mL of base.

�� We need to determine the moles of H⁺ remaining and the volume of the solution to calculate [H⁺].

�� moles = [concentration] x volume

�� moles OH^- =� (0.1 mol/L) x (0.010 L) = 0.0010 mol OH^-

�� starting moles H⁺ = (0.1 mol/L) x (0.025 L) = 0.0025 mol H⁺

�� moles H⁺ remaining after addition of 10 mL base = 0.0025 � 0.0010 = 0.0015 mol H⁺

�� volume after addition of 10.0 mL base = 25.0 mL acid + 10.0 mL base = 35 mL = 0.035 L

�� [H⁺] = 0.0015 mol / 0.035 L = 4.28 x 10^-2 M H⁺

�� pH = -log (4.28 x 10^-2) = 1.37

3.�� The pH after addition of 20 mL of base.

� ��

�� The same steps are followed, as outlined above.

�� moles OH^- = 0.0020

�� moles H⁺ remaining = 0.0025 � 0.0020 = 0.0005

�� volume = 25.0 mL + 20.0 mL = 45.0 mL

�� [H⁺] = 0.0005 / 0.045 L = .0111 M

�� pH = 1.95

4.� �� pH at the equivalence point

�� moles OH^- = 0.025

�� moles H⁺ remaining = 0.0025 � 0.0025 = 0

�� [H⁺] = [ OH^- ]�� pH = 7.00

�� There is a large change in the pH at the equivalence point.� The pH jumps from about 2 to 7.�

5.�� pH after addition of 35.0 mL of base

��

�� After the equivalence point, there will be an excess of base, which will determine the pH.

�� moles OH^- added = 0.0035 moles

�� moles H⁺ reacting with OH^‑ = 0.0025

�� moles OH^- remaining � 0.0035 � 0.0025 = 0.0010

�� volume = 25.0 mL + 35.0 mL = 60.0 mL

�� [ OH^- ] = 0.0010 mol / 0.060 L = 0.01667 M

�� pOH = - log (1.67 x 10^-2) = 1.78�� pH = 12.22

�� Again, there is a large jump in the pH above the equivalence point (pH = 7.00 to 12.22)

II. �� A weak acid with a strong base��

We will calculate the pH of 25 mL of 0.1 M CH₃COOH� titrated with 0.1 M NaOH.� At each point in the titration curve, we will need to determine two quantities, the concentration of H⁺ remaining in the solution, and the volume of the solution.� From these, we can calculate the [H⁺] and, from that, the pH.� This is a weak acid with a K _a = 1.8 x 10^-5.

1.� �� The initial pH, before the addition of strong base.�

�� In dealing with weak acids and bases, there will be an incomplete dissociation in water.� The degree of dissociation will be determined by K _a , the acid dissociation constant in this case.

�� CH₃COOH D CH₃COO^- + H⁺

�� K _a = �� [H⁺] [CH₃COO^-]

�� [CH₃COOH]

�� In this example, we have a 0.1 M solution of CH₃COOH, so the [H⁺] can be calculated using the equilibrium expression:

�� K _a = 1.8 x 10^-5 =�� [x] [x] ��, where x is the amount of weak acid that dissociates.�

�� [0.1-x]

�� x = 1.34 x 10^-3 � = [H⁺]�� pH = 2.87

�� The pH is higher in this problem than in the strong acid titration because we have partial dissociation of the acid, and

�� consequently, a lower [H⁺].

2.�� The pH after addition of 10 mL of base.

�� We need to determine the moles of CH₃COOH remaining and the volume of the solution to calculate [H⁺].

�� moles = [concentration] x volume

�� moles OH^- =� (0.1 mol/L) x (0.010 L) = 0.0010 mol OH^-

�� starting moles CH₃COOH = (0.1 mol/L) x (0.025 L) = 0.0025 mol CH₃COOH

�� moles CH₃COOH remaining after addition of 10 mL base = 0.0025 � 0.0010 = 0.0015 mol CH₃COOH

�� volume after addition of 10.0 mL base = 25.0 mL acid + 10.0 mL base = 35 mL = 0.035 L

�� Again, we will use the equilibrium expression to calculate the [H⁺].� The added base will neutralize some of the of acid,

�� and produce the conjugate base, CH₃COO^- and H₂O.� At this point in the curve, we have neutralized 0.0010 moles of the acid and produced 0.0010 moles of CH₃COO^-.� These are the new starting concentrations for the equilibrium process.� A reaction table �� (ICE) can be set up to calculate the new equilibrium concentrations.

�� [CH₃COOH] (M)�� [CH₃COO^-] (M)�� [H⁺] (M)

�� Initial�� 0.0015mol/35 mL�� 0.0010 mol/35 mL�� 0.0 mol/35 mL

�� Change�� x� �� +x�� +x

�� Equilibrium�� 0.043 � x�� 0.029 + x�� x

�� K _a = 1.8 x 10^-5 = (0.029 + x) (x) �� assume that x is << 0.043 or 0.029�� 1.8 x 10^-5 = (0.029)(x) �� x = 2.67 x 10^-5

�� (0.043 � x)�� (0.043)

��

�� x = [H⁺] �� pH = - log (2.67 x 10^-5) = 4.57

��

�� An alternative approach is to realize that these are buffering conditions, with the weak acid, CH₃COOH, and its conjugate base, �� CH₃COO^-, both present in the solution.� These conditions are appropriate for use of the Henderson-Hasselbalch Equation.

�� pH = pK _a + log( [A^-]/[HA])

�� At this point in the titration, [A^-] = 0.029 M and [HA] = 0.043 M.� The pK _a = - log (1.6 x 10^-5) = 4.75.

�� pH = 4.75 + log (0.029/0.043) = 4.75 -.17 = 4.57

�� Comparing this to the pH value of the solution of strong acid after addition of 10 mL of strong base (pH = 1.37) illustrates the �� common ion effect.� The presence of one of the products of the dissociation reaction, CH₃COO^-, lessens the extent of �� dissociation, and leads to a much higher pH, since fewer H⁺ ions are formed.

3.�� The pH after addition of 20 mL of base.

� ��

�� The same steps are followed, as outlined above.

�� moles OH^- = 0.0020 = moles CH₃COO^-

�� moles CH₃COOH remaining = 0.0025 � 0.0020 = 0.0005

�� volume = 25.0 mL + 20.0 mL = 45.0 mL

�� pH = pK _a + log ([A^-]/[HA])�� pH = 4.75 + log (0.044/0.11) = 5.35

4.� �� pH at the equivalence point

�� moles OH^- = 0.0025 = moles CH₃COO^-

�� moles CH₃COOH remaining = 0.0025 � 0.0025 = 0

�� At this point, the only species present is CH₃COO^-, the conjugate base of a weak acid.� It is a hydrolyzing species, which will �� react with water to make a basic solution.

�� CH₃COO^- + H₂O D CH₃COOH + OH^- �� K _b = K _w / K _a = 1 x 10^-14/ 1.8 x 10^-5 = 5.56 x 10^-10

�� [CH₃COO^-]�� [CH₃COOH]�� [ OH^- ]

�� Initial�� 0.0025 mol/ 50 mL�� 0.0 mol�� 0.0 mol

�� Change�� x �� + x�� + x

�� Equilibrium�� 0.05 � x �� x�� x

�� 5.56 x 10^-10 = ��x² �� x = 5.27 x 10^-6 �� pOH = 5.28�� pH = 14.00 � 5.28 = 8.72 �

�� (0.05 � x)

�� At the equivalence point, we will have a basic solution in the titration of a weak acid with a strong base.�� In the strong acid/strong �� base titration, we had a neutral solution at the equivalence point.

5.�� pH after addition of 35.0 mL of base

��

�� After the equivalence point, there will be an excess of base, which will determine the pH.

�� moles OH^- added = 0.0035 moles

�� moles CH₃COOH reacting with OH^‑ = 0.0025

�� moles OH^- remaining � 0.0035 � 0.0025 = 0.0010

�� volume = 25.0 mL + 35.0 mL = 60.0 mL

�� [ OH^- ] = 0.0010 mol / 0.060 L = 0.01667 M

�� There will be an equilibrium between the added OH^- and the CH₃COO^-, but the added base will predominate and will determine

�� the pH of the solution, as it did in the strong acid/strong base titration.

�� pOH = - log (1.67 x 10^-2) = 1.78�� pH = 12.22

�� Shown below is the plot of this titration.� At the half-way point, [CH₃COOH] = [CH₃COO^-], the pH = pK _a.� In this titration the half-�� way point comes after addition of 12.5 mL of base.� The buffering region extends between � 1 pH unit from the pK _a.� In this case, from pH = 3.75 � 5.75.

II. �� A weak base with a strong acid��

We will calculate the pH of 25 mL of 0.1 M NH₃ titrated with 0.1 M HCl.� At each point in the titration curve, we will need to determine two quantities, the concentration of H⁺ remaining in the solution, and the volume of the solution.� From these, we can calculate the [H⁺] and, from that, the pH.� This is a weak base with a K _b = 1.8 x 10^-5.

1.� �� The initial pH, before the addition of strong acid.�

�� In dealing with weak acids and bases, there will be an incomplete dissociation in water.� The degree of dissociation will be determined by K _b , the base dissociation constant in this case.

�� NH₃ + H₂O D NH₄ ⁺ + OH^-

�� K _b = �� [NH₄ ⁺] [ OH^- ]

�� [NH₃]

�� In this example, we have a 0.1 M solution of NH₃, so the [ OH^- ] can be calculated using the equilibrium expression:

�� K _b = 1.8 x 10^-5 =�� [x] [x] ��, where x is the amount of weak base that dissociates.�

�� [0.1-x]

�� x = 1.34 x 10^-3 = [ OH^- ]�� pOH = 2.87�� pH = 14.00 � 2.87 = 11.13

2.�� The pH after addition of 10 mL of acid.

�� We need to determine the moles of NH₃ remaining and the volume of the solution to calculate [H⁺].

�� moles = [concentration] x volume

�� moles H⁺ =� (0.1 mol/L) x (0.010 L) = 0.0010 mol H⁺

�� starting moles NH₃ = (0.1 mol/L) x (0.025 L) = 0.0025 mol NH₃

�� moles NH₃ remaining after addition of 10 mL base = 0.0025 � 0.0010 = 0.0015 mol NH₃

�� volume after addition of 10.0 mL acid = 25.0 mL base + 10.0 mL acid = 35 mL = 0.035 L

�� Again, we will use the equilibrium expression to calculate the [H⁺].� The added acid will neutralize some of the base,

�� and produce the conjugate acid, NH₄ ⁺ and OH^- .� At this point in the curve, we have neutralized 0.0010 moles of the base and �� produced 0.0010 moles of NH₄ ⁺.� These are the new starting concentrations for the equilibrium process.� A reaction table �� (ICE) can be set up to calculate the new equilibrium concentrations.

�� [NH₃] (M)�� [NH₄ ⁺] (M)�� [ OH^- ] (M)

�� Initial�� 0.0015mol/35 mL�� 0.0010 mol/35 mL�� 0.0 mol/35 mL

�� Change�� x� �� +x�� +x

�� Equilibrium�� 0.043 � x�� 0.029 + x�� x

�� K _b = 1.8 x 10^-5 = (0.029 + x) (x) �� assume that x is << 0.043 or 0.029�� 1.8 x 10^-5 = (0.029)(x) �� x = 2.67 x 10^-5

�� (0.043 � x)�� (0.043)

��

�� x = [ OH^- ] �� pOH = - log (2.67 x 10^-5) = 4.57�� pH = 14.00 � 4.57 = 9.43

��

�� An alternative approach is to realize that these are buffering conditions, with the weak base, NH₃, and its conjugate acid, �� NH₄ ⁺, both present in the solution.� These conditions are appropriate for use of the Henderson-Hasselbalch Equation.

�� pOH = pK _b + log( [BH⁺]/[B])

�� At this point in the titration, [BH⁺] = 0.029 M and [B] = 0.043 M.� The pK _b = - log (1.6 x 10^-5) = 4.75.

�� pOH = 4.75 + log (0.029/0.043) = 4.75 -.17 = 4.57�� pH = 9.43

3.�� The pH after addition of 20 mL of base.

� ��

�� The same steps are followed, as outlined above.

�� moles H⁺ = 0.0020 = moles NH₄ ⁺

�� moles NH₃ remaining = 0.0025 � 0.0020 = 0.0005

�� volume = 25.0 mL + 20.0 mL = 45.0 mL

�� pOH = pK _b + log ([BH⁺]/[B])�� pOH = 4.75 + log (0.044/0.11) = 5.35 pH = 8.65

4.� �� pH at the equivalence point

�� moles H⁺ = 0.0025 = moles NH₄ ⁺

�� moles NH₃ remaining = 0.0025 � 0.0025 = 0

�� At this point, the only species present is NH₄ ⁺, the conjugate acid of a weak base.� It is a hydrolyzing species, which will �� react with water to make an acidic solution.

�� NH₄ ⁺ + H₂O D NH₃ + H₃O⁺ �� K _a = K _w / K _b = 1 x 10^-14/ 1.8 x 10^-5 = 5.56 x 10^-10

�� [NH₄ ⁺]�� [NH₃]�� [H⁺]

�� Initial�� 0.0025 mol/ 50 mL�� 0.0 mol�� 0.0 mol

�� Change�� x �� + x�� + x

�� Equilibrium�� 0.05 � x �� x�� x

�� 5.56 x 10^-10 = ��x² �� x = 5.27 x 10^-6 �� pH = 5.28 ��

�� (0.05 � x)

�� At the equivalence point, we will have an acidic solution in the titration of a weak base with a strong acid.��

5.�� pH after addition of 35.0 mL of base

��

�� After the equivalence point, there will be an excess of acid, which will determine the pH.

�� moles H⁺ added = 0.0035 moles

�� moles NH₃ reacting with H⁺ = 0.0025

�� moles H⁺ remaining = 0.0035 � 0.0025 = 0.0010

�� volume = 25.0 mL + 35.0 mL = 60.0 mL

�� [H⁺] = 0.0010 mol / 0.060 L = 0.01667 M

�� There will be an equilibrium between the added H⁺ and the NH₃, but the added acid will predominate.

�� pH = - log (1.67 x 10^-2) = 1.78

�� Shown below is the plot of this titration.� At the half-way point, [NH₃] = [NH₄ ⁺], the pH = pK _a= 9.25.�� In this titration the half-�� way point comes after addition of 12.5 mL of acid.�

�� The effective buffering region extends from about pH 8.25 � 10.25.�

��

IV. Titration of a polyprotic acid

�� Polyprotic acids contain more than one acidic proton, and more than one pK _a.� An example is sulfurous acid, H₂SO₃.

�� H₂SO₃ D H⁺ + HSO₃ ^- �� K _a1 = 1.4 x 10^-2

�� HSO₃ ^- D H⁺ + SO₃ ^2- �� K _a2 = 6.5 x 10^-8

��

�� It will require 2 equivalents of base to neutralize one equivalent of the acid.�

�� We can approximate the titration curve of 40 mL of 0.10 M sulfurous acid with 0.10 M NaOH fairly easily, by making some �� simplifying assumptions.

�� 1.� The pH of the acid in water will be determined solely by the K _a1.�

��

�� In this example, the initial pH is given by:

�� 1.4 x 10^-2 = [H⁺] [HSO₃ ^-] =�� x² �� x = [H⁺] = 0.031

�� [H₂SO₃]�� (0.1 � x)

��

�� [Note:� In this case the 5% approximation is not good, and you have to solve for x either using the quadratic equation or the method �� of successive approximations.]

�� pH = - log 0.031 = 1.51

�� 2.� The pH at the first half-way point is equal to � logK _a1 = 1.85

�� 3.� The pH at the second half-way point is equal to � log K _a2 = 7.19

�� 4. � The pH at the equivalence point will be approximately equal to the average of these two pH values, (1.85 + 7.19) � 2 = 4.52.

�� 5.� The pH at the second equivalence point will be determined by the concentration of SO₃ ^2-, the conjugate base of the weak acid, HSO₃ ^-.�

�� The approximate titration curve for 0.1 M sulfurous acid is shown below.� Notice the x-axis is expressed as equivalents of base �� added.� We start with (0.1 mol / L x 0.040 L) = 0.004 moles of acid.� Each equivalent is 0.004 moles of base, or 40 mL of base.� �� There are buffering regions from pH = 0.85 to pH = 2.85 and from pH = 6.19 to pH = 8.19.

��

��

Titrations of polyprotic acids are very important in biochemistry, because proteins are comprised of amino acids, chemical species with up to three acidic protons.� The charge on the amino acids will change with the ambient pH, and their biological properties are strongly influenced by their charges.�

Find the Ph of a 0.120 M Hf Solution

Source: http://butane.chem.uiuc.edu/cyerkes/chem102aefa07/lecture_notes_102/lecture27-102.htm

Octoman Cleaskulty

Find the Ph of a 0.120 M Hf Solution

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