Whenever a **WEAK BASE** reacts with a **STRONG ACID** 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 acid remain then we will use its concentration to find the pH of the solution.

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

Example #1:

A buffer contains 167.2 mL of 0.25 M propanoic acid, CH_{3}CH_{2}COOH, with 138.7 mL of 0.42 M sodium propanoate, CH_{3}CH_{2}COONa. Find the pH after the addition of 150.2 mL of 0.56 M HCl. The K_{a} of CH_{3}CH_{2}COOH is 1.3 x 10^{-5}.

Example #2:
Calculate the pH of the solution that results from the mixing of 75.0 mL of 0.100 M NaC_{2}H_{3}O_{2} and 75.0 mL of 0.150 M HC_{2}H_{3}O_{2} with 0.0025 moles of HBr. K_{a} of HC_{2}H_{3}O_{2} is 1.8 x 10^{-5}.

You add a small amount of HCl to a solution of 0.20 M HBrO and 0.20 M NaBrO. What do you expect to happen?
1. The [OH−] will decrease slightly.
2. The H+ ions will react with the HBrO molecules.
3. The pH will increase slightly.
4. The Ka for HBrO will increase.

What is the pH of a 0.20 M NH 3/0.20 M NH4Cl buffer after 15 mL of 0.15 M HCl has been added to 65 mL of buffer?

A 250.0-mL solution buffer solution is 0.250 M in acetic acid and 0.250 M in sodium acetate. The acid dissociation constant of acetic acid is 1.8 x 10-5.
i. What is the initial pH of this solution?
ii. What is the pH after addition of 0.0050 mol of HCl?
iii. What is the pH after addition of 0.0050 mol of NaOH?

A 75.0-mL volume of 0.200 M NH 3 (Kb = 1.8 x 10-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 13.0-mL of HNO3.

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10-4.
A) 3.09
B) 4.11
C) 3.82
D) 3.46
E) 2.78

In order to create a buffer 7.321 g of potassium lactate is mix with 550.0 mL of 0.328 M lactic acid, HC3H5O3. What is the pH of the buffer solution after the addition of 700.0 mL of 0.100 M hydrobromic acid, HBr? The Ka of HC3H5O3 is 1.4 x 10-4.

A buffer contains 55.3 mL of 0.200 M nitrous acid, HNO 2, with 100.0 mL of 0.550 M sodium nitrite, NaNO2. Find the pH after the addition of 180.0 mL of 0.400 M HClO 4. The Ka of HNO2 is 4.6 x 10-4.

What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.50 M HCl to reach the equivalence point? The acid dissociation constant of HOCl is 3.0x10-8.
3.76
0.30
3.18
4.03

Which of the following is the best choice to titrate 0.10 M C 6H5NH2?
a) 0.30 M NaOH
b) 0.01 M NH 3
c) 0.10 M HNO3
d) 0.10 M HCO2H

Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.8 x 10 -10) with 0.500 M HCl. What is the pH of the solution at the stoichiometric point?
A. -0.85
B. 8.70
C. 2.68
D. 11.62
E. none of these

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at 40 mL of added acid.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at 50 mL of added acid.

A sample of 7.6 L of NH3 gas at 22 oC and 735 torr is bubbled into a 0.50-L solution of 0.40
M
HCl.Assuming that all the NH3 dissolves and that the volume of the solution remains 0.50 L, calculate the pH of the resulting solution.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at 10 mL of added acid.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at 20 mL of added acid.

Calculate the equilibrium concentrations of Na+, Cl-,H+, CHO2-, and HCHO2 when 50.0 mL of 0.35
M
HCl is mixed with 50.0 mL of 0.35
M
NaCHO2.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl.Calculate the pH at equivalence point.

A student creates 150.00mL of a buffer that is 0.018 M in formic acid and 0.016 M in sodium formate. What would the pH of the buffer be after the addition of 15.00 mL of a 0.070 M HCl solution to the buffer?A. 3.69B. 3.85C. 4.18D. 1.15E. 3.30

Calculate the pH during the titration of 20.00 mL of 0.1000 M ammonia with 0.1000 M HCl(aq) after 15 mL of the acid have been added. Kb of ammonia = 1.8 x 10-5.

How does a buffer resist change in pH upon addition of a strong acid?a) The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.b) The strong acid reacts with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH.c) The strong acid reacts with the strong base in the buffer to form a salt which, produces few H + ions in solution and therefore only a little change in pH.

What is the pH of a 0.20 M NH 3/0.20 M NH4Cl buffer after 10 mL of 0.10 M HCl has been added to 65 mL of buffer?

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 MHCl. Calculate the pH at equivalence point. Express your answer using two decimal places.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100M HCl. Calculate the pH at 50 mL of added acid. Express your answer using two decimal places.

Which of the following occurs when HCl is added to a buffer containing (CH3)3N and (CH3)3NH+? a. The concentration of (CH3)3N will increase.b. The concentration of (CH3)3NH+ will increase.c. The concentration of (CH3)3N will decrease.d. The concentration of (CH3)3NH+ will decrease.

20 mL of 0.25 M of NH 3 is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. Kb (NH3) = 1.8 × 10 −5.a. 12.50b. 7.00c. 4.74d. 1.12

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant:(g) 20.05 mL

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant:(h) 25.00 mL

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.72.0 mL

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.73.3 mL

Consider the titration of 100.0 mL of 0.100 M H 2NNH2 (Kb = 3.0 x 10 -6) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO 3 have been added.f. 100.0 mL

Consider the titration of a 26.0-mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following.the pH after adding 6.0 mL of acid beyond 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.)b. Calculate the pH after 50.0 mL of KOH has been added.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl. Calculate the pH at each volume of added acid: 0 mL, 10 mL, 20 mL, equivalence point, one-half equivalence point, 40 mL, 50 mL. Sketch the titration curve.

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)e. Calculate the pH after 300.0 mL of HCl has been added.

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)f. At what volume of HCl added does the pH = 6.04?

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant:(a) 0 mL

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl 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 triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant:(c) 15.00 mL

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl 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 triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant:(e) 19.95 mL

A 20.0 mL sample of 0.150 M ethylamine is titrated with 0.0981 M HCl. What is the pH after the addition of 5.0 mL of HCl? For ethylamine, pKb= 3.25.

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.0 mL

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.20.0 mL

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.59.1 mL

Consider the titration of 100.0 mL of 0.100 M H 2NNH2 (Kb = 3.0 x 10 -6) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO 3 have been added.a. 0.0 mL

Consider the titration of 100.0 mL of 0.100 M H 2NNH2 (Kb = 3.0 x 10 -6) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO 3 have been added.b. 20.0 mL

Consider the titration of 100.0 mL of 0.100 M H 2NNH2 (Kb = 3.0 x 10 -6) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO 3 have been added.c. 25.0 mL

Consider the titration of 100.0 mL of 0.100 M H 2NNH2 (Kb = 3.0 x 10 -6) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO 3 have been added.d. 40.0 mL

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 HCl for the titration of 25.0 mL of 0.100 M NH3 (Kb = 1.8 x 10-5) with 0.100 M HCl. Plot the results of your calculations as pH versus milliliters of HCl 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, 25.0 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the hydrochloric acid for the titration of 25.0 mL of 0.100 M pyridine with 0.100 M hydrochloric acid (Kb for pyridine is 1.7 x 10 -9). Plot the results of your calculations as pH versus milliliters of hydrochloric acid added.

In the titration of 50.0 mL of 1.0 M methylamine, CH3NH2 (Kb = 4.4 x 10-4), with 0.50 M HCl, calculate the pH under the following conditions.a. after 50.0 mL of 0.50 M HCl has been added

Consider the titration of a 26.0-mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following.the pH at 6.0 mL of added acid

Consider the titration of a 26.0-mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following.the pH at one-half of the equivalence point

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl. Calculate the pH at one-half equivalence point.

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)a. Calculate the pH after 0.0 mL of HCl has been added.

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)b. Calculate the pH after 25.0 mL of HCl has been added.

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)c. Calculate the pH after 70.0 mL of HCl has been added.

Consider the following four titrations (i–iv):i. 150 mL of 0.2 M NH 3 (Kb = 1.8 x 10 -5) by 0.2 M HClii. 150 mL of 0.2 M HCl by 0.2 M NaOHiii. 150 mL of 0.2 M HOCl (K a = 3.5 x 10 -8) by 0.2 M NaOHiv. 150 mL of 0.2 M HF (K a = 7.2 x 10 -4) by 0.2 M NaOHa. Rank the four titrations in order of increasing pH at the halfway point to equivalence (lowest to highest pH).

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH 3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant:(f) 20.00 mL

Find the pH of the equivalence point(s) and the volume (mL) of 0.125 M HCl needed to reach the point(s) in titrations of(a) 65.5 mL of 0.234 M NH 3

Find the pH of the equivalence point(s) and the volume (mL) of 0.125 M HCl needed to reach the point(s) in titrations of(b) 21.8 mL of 1.11 M CH 3NH2

Find the pH of the equivalence point(s) and the volume (mL) of 0.447 M HNO 3 needed to reach the point(s) in titrations of(a) 2.65 L of 0.0750 M pyridine (C5H5N)

Find the pH of the equivalence point(s) and the volume (mL) of 0.447 M HNO 3 needed to reach the point(s) in titrations of(b) 0.188 L of 0.250 M ethylenediamine (H 2NCH2CH2NH2)

Assume that 31.0
mL of a 0.10 M solution of a weak base B
that accepts one proton is titrated with a 0.10 M solution of the
monoprotic strong acid HA.How many moles of HA have been added at the equivalence point?

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.60.0 mL

Consider the titration of 100.0 mL of 0.100 M H 2NNH2 (Kb = 3.0 x 10 -6) by 0.200 M HNO3. Calculate the pH of the resulting solution after the following volumes of HNO 3 have been added.e. 50.0 mL

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.b. 100.0 mL of 0.10 M C2H5NH2 (Kb = 5.6 x 10 -4) titrated by 0.20 M HNO3

In the titration of 50.0 mL of 1.0 M methylamine, CH3NH2 (Kb = 4.4 x 10-4), with 0.50 M HCl, calculate the pH under the following conditions.b. at the stoichiometric point

Consider the titration of a 26.0-mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following.the pH at the equivalence point

Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCl. (K b for HONH2 = 1.1 x 10-8.)d. Calculate the pH at the equivalence point.

Consider the following four titrations (i–iv):i. 150 mL of 0.2 M NH 3 (Kb = 1.8 x 10 -5) by 0.2 M HClii. 150 mL of 0.2 M HCl by 0.2 M NaOHiii. 150 mL of 0.2 M HOCl (K a = 3.5 x 10 -8) by 0.2 M NaOHiv. 150 mL of 0.2 M HF (K a = 7.2 x 10 -4) by 0.2 M NaOHb. Rank the four titrations in order of increasing pH at the equivalence point.

Consider the following four titrations (i–iv):i. 150 mL of 0.2 M NH 3 (Kb = 1.8 x 10 -5) by 0.2 M HClii. 150 mL of 0.2 M HCl by 0.2 M NaOHiii. 150 mL of 0.2 M HOCl (K a = 3.5 x 10 -8) by 0.2 M NaOHiv. 150 mL of 0.2 M HF (K a = 7.2 x 10 -4) by 0.2 M NaOHc. Which titration requires the largest volume of titrant (HCl or NaOH) to reach the equivalence point?

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.00 mL sample of the cleaner, the equivalence point is reached after 16.85 mL of 0.138 M HCl has been added. What the initial concentration of ammonia in the solution? What is the pH of the solution at the equivalence point?

A 1.00 L buffer solution is 0.150 M in HC 7H5O2 and 0.250 M in LiC7H5O2. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HC7H5O2 is 6.5 x 10-5.