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Henderson Hasselbalch: Titrations

Related Topics

Wize University Chemistry Textbook > Buffers and Titrations

Indicators

2 Activities

Wize University Chemistry Textbook > Chemical Reactions

Titrations

2 Activities

Wize University Chemistry Textbook > Buffers and Titrations

Titration Problems

2 Activities

Suppose you performed a potentiometric titration on the solutions in flasks A, B, and C by a gradual addition of HCℓ (aq) to each flask shown.


Choose one option per statement to describe the differences between these titration curves.

When the solution in flask B is titrated…

(B1) … the initial pH will be lower / higher / the same as the initial pH in flask A.
(B2) …the equivalence point will be reached after adding a smaller / larger / equal volume of HCℓ(aq) as was needed to reach equivalence in titration A.
(B3) …the equivalence point will occur at lower / higher / the same pH vs. titration A.
(B4) …the half-equivalence point will be reached after adding a smaller / larger / equal
volume of HCℓ(aq) as was needed to reach half-equivalence in titration A.
(B5) …the half-equivalence point will occur at lower / higher / the same pH vs. titration A.

When the solution in flask C is titrated…

(C1) … the initial pH will be lower / higher / the same as the initial pH in flask A.
(C2) …the equivalence point will be reached after adding a smaller / larger / equal volume of HCℓ(aq) as was needed to reach equivalence in titration A.
More Indicators Questions:
The indicator cresol red has Ka = 5.0 x 10-9. Over what approximate pH range does it change colour?
Phenolphthlein is used as an indicator in acid base titrations because it is halochromic, it changes color based on pH. Shown below are the two forms of phenolphthalein at low pH and high pH. One of the forms is colorless and one is bright purple. Which one is colored? Explain your answer.
Henderson Hasselbalch: Buffers and Indicators
The indicator cresol red has Ka = 5.0 x 10-9. Over what approximate pH range does it change colour?
Sketch the titration curve for a weak acid (Ka = 3.4 x 10-6) being titrated by a strong base. What is the pH at the equivalence point? What would be the ideal pKa value for an indicator in this titration?
For which type of titration will the pH be basic at the equivalence point?
More Titrations Questions:
Titrations
A student prepares a potassium hydrogen phosphate (K2HPO4, MW = 156.1739 g/mol) solution and pipettes 20.00 mL of this solution for a titration. In the titration, it required 17.51 mL of 0.09977 M HCl to reach the endpoint. What is the concentration of the potassium hydrogen phosphate solution?
25L of a solution of 0.5M KOH is required to neutralize 15L of sulphuric acid. What is the concentration of the acid?
KOH + H2SO4 --> K2SO4 + 2H2O
More Titration Problems Questions:
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Acids and Bases: Titrations
25mL of a solution of 0.5mol/L KOH is required to neutralize 15mL of sulphuric acid. What is the concentration of the acid?
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(ℓ)2KOH\left(aq\right)+H_2SO_4(aq)\to K_2SO_4(aq)+2H_2O(\ell)
Henderson Hasselbalch: Titrations and pH
For the titration of 50.00 mL of 0.1000 M ammonia (NH3) with 0.1000 M HCl calculate the pH (For ammonia Kb​=1.8×10−5)(For\ ammonia\ Kb​=1.8\times10^{-5})
Before the addition of any HCl
After 20.00 mL of the acid has been added
Redox
A 5.000 mL sample of a solution of hydrogen peroxide, H2O2, was analyzed by titration with potassium permanganate, KMnO4, according to the reaction shown below. The H2O2 sample required 42.8 mL of 0.0175 mol L−1 KMnO4 (aq) to reach the end point.
2 KMnO4 + 5 H2O2 + 6 H+ → 5 O2 + 2 Mn2+ + 2 K+ + 8 H2O
What was the concentration of the H2O2 solution? Choose the closest value.
To a 100 mL of 0.1M of CH3COOH\rm CH_3COOH (acetic acid) is added 25mL of 0.17M NaOH\rm NaOH. The Ka\rm K_a of acetic acid is 1.8×10−51.8 \times 10^{-5}. Calculate the pH\rm pH of the solution. Please report your answer to two decimal places.
25L of a solution of 0.5M KOH is required to neutralize 15L of sulphuric acid. What is the concentration of the acid? Balance the equation if it is unbalanced.
KOH + H2SO4 --> K2SO4 + 2H2O
What volume of hydrochloric acid of 0.1M is required to neutralize 500mL at 0.5M sodium hydroxide?
HCl + NaOH --> NaCl + H2O
Titration Problem
We titrated 40mL of 0.2M HBr with 0.4M NaOH...
The pH meter indicated that the initial pH was 2.9, pH at the half equivalence point was 3.5, equivalence pt was at pH 8.9.
a) What is Ka for HBr?
What volume of 3.16 M H3PO4 solution is necessary to react with 27 mL of 0.18 M Mg(OH)2?
Acid Base Calculation
An impure sample of solid Na2CO3 was reacted with 0.1755 M HCl. A 0.2337 g solid sample required 15.55 mL of HCl solution.
Practice: Long Titration Problem
For the titration of 50.00 mL of 0.1000 M ammonia (NH3) with 0.1000 M HCl calculate the pH
Before the addition of any HCl
After 20.00 mL of the acid has been added
Acids and Bases: Strong Acid Mixture
Practice Question: Strong Base Mixture
Calculate the pH and pOH of a solution containing a mixture of 75.00 mL of 0.100 M HI and 25.00 mL of 0.250 M HClO4.
Buffers: Titration of Sodium Citrate
A student dissolves a sodium citrate (Na3C6H5O7; MW = 258.069 g/mol) tablet in 350.00 mL of water. The student then pipetted 20.00 mL of the solution and performed a titration using a stock solution of HCl previously standardized to a concentration of 0.9899 M. The experiment required 33.55 mL of the HCl solution to reach the endpoint and fully neutralize the sodium citrate solution. Calculate the concentration of the sodium citrate solution.
Titrations
A student prepares a potassium hydrogen phosphate (K2HPO4, MW = 156.1739 g/mol) solution and pipettes 20.00 mL of this solution for a titration. In the titration, it required 23.43 mL of 0.100 M HCl to reach the endpoint. What is the concentration of the potassium hydrogen phosphate solution?
Strong Acids and Bases
Calculate the pH of a basic solution prepared by taking 10.00 mL of 0.700 M NaOH solution and diluting it to 100.00 mL with water.
Titrations
A student prepares a potassium hydrogen phosphate (K2HPO4, MW = 156.1739 g/mol) solution and pipettes 20.00 mL of this solution for a titration. In the titration, it required 17.51 mL of 0.09977 M HCl to reach the endpoint. What is the concentration of the potassium hydrogen phosphate solution?
pH: Neutralization
A student adds 11.6 mL of a 3.7M KOH solution, to 29 mL of a 1.4 M HCl. What is the final pH after the addition?
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution
Chemical Reactions: Limiting Reagent
15mL of a 1.2M solution of HCl reacts with 22mL of a 1.4M solution of NaOH. How many moles of NaOH are left after the reaction is complete?
HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)HCl_{(aq)}+NaOH_{(aq)}\rightarrow H_2O_{(l)}+ NaCl_{(aq)}
For the titration of 50.00 mL of 0.1000 M ammonia (NH3) with 0.1000 M HCl calculate the pH. Enter 2 decimal places with all answers. Kb=1.8x10-5
Before the addition of any HCl
After 20.00 mL of the acid has been added
Acids and Bases: Titrations
25L of a solution of 0.5M KOH is required to neutralize 15L of sulphuric acid. What is the concentration of the acid? Balance the equation if it is unbalanced.
KOH + H2SO4 --> K2SO4 + 2H2O
Acids and Bases: Titrations
What volume of hydrochloric acid of 0.1M is required to neutralize 500mL at 0.5M sodium hydroxide?
HCl + NaOH --> NaCl + H2O
For which type of titration will the pH be basic at the equivalence point?
A student titrated 10.00 mL of 0.10 M sodium acetate (CH3COONa) with HCℓ and determined that 10.00 mL of HCℓ was needed to reach the equivalence point. At equivalence, what is the expected pH?
Ka of CH3COOH = 1.8×10-5
A student titrated 10.00 mL of 0.10 M sodium acetate (CH3COONa) with HCℓ and determined that 10.00 mL of HCℓ was needed to reach the equivalence point. At equivalence, what is/are the major species in solution (besides water and any pH-neutral ions)?
Calculate the pH of the solution resulting from the addition of 10.0 mL of a 0.10 M NaOH to 50.0 mL of 0.10 M HCN (Ka = 4.9 x 10-10) solution