Whenever a **WEAK ACID** reacts with a **STRONG BASE** we use an **ICF Chart** to determine the pH of the solution.

Concept #1: Understanding Weak Acid–Strong Base titration reactions

Concept #2: If you use an ICF Chart and at the end you have remaining weak acid and conjugate base then you have a buffer so you use the **Henderson-Hasselbalch Equation **to find pH.

Concept #3: If any excess moles of the strong base remain then we will use its concentration to find the pOH then pH of the solution.

Concept #4: If the moles of both the weak acid and strong base are equal then we will have only conjugate base at the end of our ICF Chart Calculation. To find pH we would follow up with an ICE Chart.

Example #1:

Consider the titration of 75.0 mL of 0.0300 M H_{3}C_{3}O_{3} (K_{a} = 4.1 X 10^{-3}) with 12.0 mL of 0.0450 M KOH. Calculate the pH.

Example #2:

In order to create a buffer, 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 175.0 mL of 0.300 M NaH?

Example #3:

Consider the titration of 75.0 mL of 0.60 M HNO_{2} with 0.100 M NaOH at the equivalence point. What would be the pH of the solution at the equivalence point? The K_{a} of HNO_{2} is 4.6 x 10^{-4}.

Determine the pH of a solution created by mixing 127.6 mL of 0.278 M formic acid, HCO2H, with 148.3 mL of 0.329 M sodium hydroxide, NaOH.
A. 1.32
B. 10.35
C. 3.60
D. 12.68
E. 4.17

1.0-L of a buffer solution contains 0.800 moles of acid HA (K a = 2.00 x 10 -6 ) and 0.800 moles of its conjugate base, A− . What is the change in pH upon the addition of 0.0500 moles of a strong base to this solution (with no change in volume)?
(a) +0.05
(b) +0.03
(c) +0.01
(d) -0.01
(e) -0.05

In a titration, 25.00 mL of aqueous hydrofluoric acid requires the addition of 30.00 mL of 0.400 M NaOH to reach the equivalence (stoichiometric) point. What will be the corresponding pH? Ka (HF) = 6.76 × 10 -4 .
(a) 9.12
(b) 8.25
(c) 5.75
(d) 7.00
(e) 1.74

Consider a buffer composed of a mixture of acetic acid and sodium acetate in equal amounts. Which of the following clearly indicates what happens to the buffer when a very small amount of concentrated LiOH is added to it.
[CH3COOH] [CH 3COO – ] pH
A. increases decreases decreases
B. decreases increases increases
C. increases decreases increases
D. decreases increases decreases
E. increases increases increases

What is the pH of a 1.00 L solution that is made by starting with 0.100 M HCOOH
(Ka = 1.8 X 10−4) with 0.035 moles of NaOH added?
A. 3.29 B. 3.48 C. 3.75 D. 4.00 E. 12.54

A solution is made by mixing 500 mL of 0.167 M NaOH with 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentrations of H +, CH3COOH, CH3COO- , OH-, and Na+.

What is the pH of a one-liter solution that is 0.100 M in NH 3 and 0.100 M in NH 4Cl after 1.6 g of NaOH(s) has been added? Kb for NH3 is 1.8 × 10-5.
(a) 4.74
(b) 9.26
(c) 9.46
(d) 9.62
(e) 11.12

A 25.0 mL sample of 0.150 M hydrofluoric acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of hydrofluoric acid is 3.5 × 10-4.
(a) 3.45
(b) 7.00
(c) 8.17
(d) 10.17
(e) 10.83

What is the pH of a solution that contains 1.0 L of 0.10 M CH3COOH and 0.080 M NaCH3COO after 0.03 moles of NaOH added?
A. 4.33
B. 4.65
C. 4.74
D. 4.94
E. 12.48

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The K a of HF is 3.5 × 10 -4.
A) 2.08
B) 3.15
C) 4.33
D) 3.46
E) 4.15

A formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer?
a) 3.67
b) 3.78
c) 3.81
d) 3.85
e) 3.95

Find the pH of a solution made by titrating 60.0 mL of 0.125 M HCO 2H with 50.0 mL of 0.100 M LiH. The Ka of HCO2H is 1.8 x 10-4.

Calculate the pH of the solution resulting from the mixing of 55.0 mL of 0.100 M NaCN and 75.0 mL of 0.100 M HCN with 0.0090 moles of NaOH. The Ka of HCN is 4.9 x 10-10.

A solution is prepared by dissolving 0.23 moles of hydrazoic acid 0.27 moles of sodium azide in water sufficient to yield 1.00L of solution. The addition of 0.05 mol of sodium hydroxide to this buffer causes the pH to increase slightly. The pH does not increase drastically because the sodium hydroxide reacts with the ______ present in the buffer solution. The Ka of hydrazoic acid is 1.9x10 –5.
Hydrazoic acid
Azide
H2O
This is a buffer solution: the pH does not change upon the addition of an acid or base.
All of the above

A 25.0mL sample of 0.150M of hydrazoic acid is titrated with 0.150M sodium hydroxide solution. What is the pH after 13.3mL of base is added? The Ka of hydrazoic acid is 1.9x10 –5.
A. 4.45
B. 3.03
C. 1.34
D. 4.66
E. 4.78

What is the pH of a solution that contains 1.0 L of 0.10 M CH 3COOH and 0.080 M NaCH3COO after 0.03 moles of NaOH added?
A. 4.33
B. 4.65
C. 4.74
D. 4.94
E. 12.48

In the titration of a weak acid with a strong base, how would you calculate these quantities?initial pH

In the titration of a weak acid with a strong base, how would you calculate these quantities?pH before the equivalence point

In the titration of a weak acid with a strong base, how would you calculate these quantities?pH at one-half the equivalence point

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at 0 mL of added base.

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at 5 mL of added base.

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at 10 mL of added base.

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at one-half of the equivalence point.

As the acid is titrated, the pH of the solution after the addition of 11.10 mL of the base is 4.88. What is the Ka for the acid?

In the titration of a weak acid with a strong base, how would you calculate these quantities?pH beyond the equivalence point

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at 25 mL of added base.

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at 20 mL of added base.

In the titration of a weak acid with a strong base, how would you calculate these quantities?pH at the equivalence point

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.Calculate the pH at the equivalence point.

What is the pH at the equivalence point when 75.00 mL of a 0.150 M solution of acetic acid (CH3COOH) is titrated with 0.10 M NaOH to its end point?

Calculate the pH of a solution after adding 28.522 mL of 0.285 M NaOH to 32.817 mL of a 0.318 M benzoic acid solution. Ka of benzoic acid is 6.5 x 10-5. A. 4.08B. 4.74C. 3.64D. 3.97E. 4.19

What would be the final pH if 0.0100 moles of solid NaOH were added to 100mL of a buffer solution containing 0.600 molar formic acid (ionization constant = 1.8x10-4 ) and 0.300 M sodium formate?3.654.053.843.35

Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of a 0.10 M solution of aspirin (acetylsalicyclic acid, Ka = 3.0 × 10–4) solution.A) 3.5 B) 2.9 C) 4.1 D) 10.5 E) 1.8

A 30.0-mL volume of 0.50 M CH 3COOH (Ka = 1.8 x 10 -5) was titrated with 0.50 M NaOH. To reach the equivalence point, 30.0-mL of NaOH must be added. What will be the pH?

What is the pH of a 500 mL buffer containing 0.10 M Na 2HPO4/0.15 M KH2PO4 after adding 19 mL of 1 M NaOH?

A certain weak acid, HA, with a Ka value of 5.61 × 10 −6, is titrated with NaOH.A solution is made by titrating 7.00 mmol (millimoles) of HA and 1.00 mmol of the strong base. What is the resulting pH?Express the pH numerically to two decimal places.

An aqueous solution contains dissolved NH4Cl and NH3. The concentration of NH3 is 0.50 M and the pH is 8.95. The Kb of NH3 is 1.8x10-5a. What is the molar concentration of NH4 + in the buffer? b. What is the pH after adding 0.10 moles of NaOH to 1.00 L of this solution? Assume no volume change upon addition of the NaOH?

A 30.0 ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH.
a. Calculate the pH at 0 mL of added base.
b. Calculate the pH at 5 mL of added base.
c. Calculate the pH at 10 mL of added base.
d. Calculate the pH at the equivalent point.
e. Calculate the pH at one-half of the equivalence point.
f. Calculate the pH at 20 mL of added base.
g. Calculate the pH at 25 mL of added base.

Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.(a) no KOH added

Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.(b) 20 mL of KOH solution added

Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.(c) 39 mL of KOH solution added

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.0 mL of the base

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.17.5 mL of the base

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.34.5 mL of the base

A 0.552-g sample of ascorbic acid (vitamin C) was dissolved in water to a total volume of 20.0 mL and titrated with 0.1103 M KOH, and the equivalence point occurred at 28.42 mL. The pH of the solution at 10.0 mL of added base was 3.72.Make a rough sketch of the titration curve from this data by calculating the pH at the beginning of the titration, at one-half of the equivalence point, at the equivalence point, and at 5.0 mL beyond the equivalence point.

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10 -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.a. 0.0 mL

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10 -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.b. 50.0 mL

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10 -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.c. 100.0 mL

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10 -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.d. 150.0 mL

Lactic acid is a common by-product of cellular respiration and is often said to cause the “burn” associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH for the titration of 25.0 mL of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 x 10 -5) with 0.100 M NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.

Consider the titration of a 23.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.the initial pH

Consider the titration of a 23.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.the pH at 6.00 mL of added base

Consider the titration of a 23.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.the pH at one-half of the equivalence point

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. Calculate the pH at each volume of added base: 0 mL, 5 mL, 10 mL, equivalence point, one-half equivalence point, 20 mL, 25 mL. Sketch the titration curve.

Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C. (K a for HCN = 6.2 x 10-10.)a. Calculate the pH after 0.0 mL of KOH has been added.

Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C. (K a for HCN = 6.2 x 10-10.)c. Calculate the pH after 75.0 mL of KOH has been added.

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (a) 0 mL

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (b) 10.00 mL

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (c) 15.00 mL

A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1532 M solution of NaOH at 25°C.(a) What is the pH of the HF solution before titrant is added?

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (d) 19.00 mL

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (e) 19.95 mL

A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1532 M solution of NaOH at 25°C.(c) What is the pH at 0.50 mL before the equivalence point?

Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.(e) 41 mL of KOH solution added

Scenes A to D represent tiny portions of 0.10 M aqueous solutions of a weak acid HA (red and blue; Ka = 4.5×10−5), its conjugate base A−(red), or a mixture of the two (only these two species are shown):(c) Arrange the scenes in sequence, assuming that they represent stages in a weak acid–strong base titration.

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.35.5 mL of the base

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.50.0 mL of the base

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10 -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.f. 250.0 mL

Consider the titration of a 23.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.the pH after adding 6.00 mL of base beyond the equivalence point

An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of(d) 1.0 L of the solution in part (a) after 0.0050 mol of NaOH has been added.

Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C. (K a for HCN = 6.2 x 10-10.)e. Calculate the pH after 125 mL of KOH has been added.

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (g) 20.05 mL

A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1532 M solution of NaOH at 25°C.(e) What is the pH at 0.50 mL after the equivalence point?

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (h) 25.00 mL

Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka = 9.8 × 10−5) with 0.100 M KOH.(d) 40 mL of KOH solution added

Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42.2 mL of 0.0520 M CH 3COOH

Find the pH of the equivalence point(s) and the volume (mL) of 0.0588 M KOH needed to reach the point(s) in titrations of(a) 23.4 mL of 0.0390 M HNO2

A 10.0 mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.0998 M NaOH.What is the pH at the equivalence point? For hydrocyanic acid, pKa = 9.31.

Determine the pH at the equivalence point for the titration of HNO2 and KOH.

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added.35.0 mL of the base

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10 -5) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.e. 200.0 mL

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.a. 100.0 mL of 0.10 M HC7H5O2 (Ka = 6.4 x 10-5) titrated by 0.10 M NaOH

Consider the titration of a 23.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.the pH at the equivalence point

Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C. (K a for HCN = 6.2 x 10-10.)d. Calculate the pH at the equivalence point.

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant: (f) 20.00 mL

A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1532 M solution of NaOH at 25°C.(d) What is the pH at the equivalence point?

A weak monoprotic acid is titrated with 0.100 M NaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. After 25.0 mL of base is added, the pH of the solution is 3.62.Estimate the pKa of the weak acid.

A 1.0L buffer solution contains 0.100 mol of HC2H3O2 and 0.100 mol of NaC2H3O. The value of Ka for HC2H3O2 is 1.8×10−5. Calculate the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer.

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO can be found here.(a) before addition of any KOH(b) after addition of 25.0 mL of KOH(c) after addition of 30.0 mL of KOH(d) after addition of 50.0 mL of KOH(e) after addition of 60.0 mL of KOH

A 250.0 mL solution of 0.100 M HClO is titrated with 0.200 M NaOH. What is the expected pH of the resulting solution once 50.0 mL of the NaOH solution has been added to the HClO solution?A. 4.68B. 7.36C. 7.71D. 8.20E. 13.30

Calculate the pH at the equivalence point from the titration of 80.00 mL of 0.140 M hydrofluoric acid, HF, with 160.00 mL of 0.0700 M NaOH. Ka of HF is 3.5 x 10-4.A. 5.94B. 8.06C. 7.75D. 8.30E. 11.61

An analytical chemist is titrating 202.5 mL of a 0.2200 M solution of propionic acid (HC2H5CO2) with a 0.5400 M solution of KOH. The pK a of propionic acid is 4.89. Calculate the pH of the acid solution after the chemist has added 89.35 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.

If 0.3705 g of pure KHP solid were NaoH and the endpoint was reached at 24.80 mL, what is the concentration of the NaOH solution? (NOte: MW of KHP = 204.223 g/mol).