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Methyl diethanolamine CH3N(C2H4OH)2 is a weak base which we will abbreviate as …

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Wize University Chemistry Textbook > Acids and Bases

pH

4 Activities

Wize University Chemistry Textbook > Acids and Bases

Acid Equations (Ka, pKa)

3 Activities

Methyl diethanolamine CH3N(C2H4OH)2 is a weak base which we will abbreviate as mdea. The conjugate acid, mdeaH+, has a pKa = 8.52.
What is the pH of a 0.15 M aqueous solution of mdea at 298K?
More pH Questions:
Weak Acids and Bases
Calculate the pH of the following solution.
A solution prepared by dissolving 4.25 x 10-4 g of CH3COONa in 100.0 mL of water.
Note: Acetic Acid Ka = 1.8 x 10-5
Weak Acids and Bases
Calculate the pH and pOH of the following solutions.
a) A 6.30 x 10-2 M solution of CH3COOH in water.
b) A solution prepared by dissolving 4.25 x 10-4 g of CH3COONa in 100.0 mL of water.
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
Molecular formula: Concentration
A solution of HBr has a pOH of 12.2, what is the [H+] concentration?
100 mL of a 0.5 M solution of nitric acid (HNO3\rm HNO_3) was added to 75 mL of a 0.37 M solution of sodium benzoate (C6H5COONa\rm C_6H_5COONa) the conjugate base of benzoic acid (C6H5COOHKa=6.6×10−5\rm C_6H_5COOH Ka = 6.6 \times 10^{-5}). Calculate the pH\rm pH of the resulting solution.
If the pH of pure water is 6.80 at 37 °C, calculate Kw pOH and [OH- ] at this temperature.
In a 0.25 M solution, a weak acid is 3.0% dissociated. Calculate [H3O+], pH and pOH of the solution
In a 0.25 M solution, a weak acid is 3.0% dissociated. Calculate [H3O+], pH and pOH of the solution
Acids and Bases: pH of Alkaline Earth Metal Hydroxides
Calculate the pH of a 1.00 x 10-4 M Mg(OH)2 solution.
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.
Weak Acids and Bases
Calculate the pH of a solution prepared by dissolving 0.500 g of sodium hydrogen phosphate (Na2HPO4, MW = 141.9579 g/mol) in 150.0 mL of water.
Weak Acids and Bases
Calculate the pH and pOH of the following solutions.
a) A 6.30 x 10-2 M solution of CH3COOH in water.
b) A solution prepared by dissolving 4.25 x 10-4 g of CH3COONa in 100.0 mL of water.
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.
Strong Acids and Bases
Calculate the pH of a solution prepared by adding 100.0 mL of 3.00 x 10-2 M KOH with 50.0 mL of 2.50 x 10-2 M Be(OH)2.
Strong Acids and Bases
Calculate the pH and pOH of the following solutions.
a) 2.75 x 10-3 M aqueous HNO3 solution.
b) 1.50 x 10-2 g of CaO dissolved in 200.0 mL of water.
Weak Acids and Bases
Calculate the pH and pOH of the following solutions, knowing that Ka = 1.8x10^-5.
a) A 5.33 x 10-2 M solution of CH3COOH in water.
b) A solution prepared by dissolving 1.15 x 10-3 g of CH3COONa in 100.0 mL of water.
Weak Acids and Bases
Calculate the pH of a solution prepared by dissolving 0.150 g of sodium hydrogen phosphate (Na2HPO4, MW = 141.9579 g/mol) in 125.0 mL of water.
Ka (HPO42−) = 4.2 x 10−13;  Kb (HPO42−) = 1.6 x 10−7. K_a\ \left(HPO_4^{2-}\right)\ =\ 4.2\ x\ 10^{-13};\ \ K_b\ \left(HPO_4^{2-}\right)\ =\ 1.6\ x\ 10^{-7}.\
Strong Acids and Bases
Calculate the pH of a basic solution prepared by taking 20.00 mL of 0.500 M Be(OH)2 solution and diluting it to 100.00 mL with water.
Strong Acids and Bases
Calculate the pH of a solution prepared by adding 25.00 mL of 0.100 M HCl with 75.00 mL of 2.50 x 10-2 M HBr.
Strong Acids and Bases
Calculate the pH and pOH of the following solutions.
a) 3.00 x 10-4 M aqueous HClO4 solution.
b) 8.00 x 10-3 g of Na2O dissolved in 150.0 mL of water.
Consider the following reaction in water:
NaOH <----> Na+ + OH-
This reaction has a Kb of 1 x 10-10 and a pOH of 3. If there is 1 M of NaOH at equilibrium, what is the concentration of Na+ in solution?
Acids and Bases: pH based on % dissociation
In a 0.25 M solution, a weak acid is 3.0% dissociated. Calculate [H3O+], pH and pOH of the solution
If an answer is an exponent such as 4.3x10-7, input the answer as: 4.3e-7
Which of the solutions below would have the HIGHEST pH value?
Acids and Bases: pH
What is the pH of a 0.10 M solution of NaF? (Ka for HF = 6.8×10-4)?
Methyl diethanolamine CH3N(C2H4OH)2 is a weak base which we will abbreviate as mdea. The conjugate acid, mdeaH+, has a pKa = 8.52.
More Acid Equations (Ka, pKa) Questions:
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What are the Ka and Kb of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
Nitrous acid has a Ka of 7.1 x 10-5. What are [H3O+], [NO2- ] andin 0.50 M HNO2?
Acids and Bases: Weak Acid Practice
Nitrous acid has a Ka of 7.1 x 10-5. What are [H3O+] and [NO2- ] in 0.50 M HNO2?
Nitrous acid has a Ka of 7.1 x 10-5. What are [H3O+], [NO2- ] andin 0.50 M HNO2?
Acids and Bases: Ka
The following occurs when HCl is placed in water:
HCl <----> H+ + Cl-
The Ka for this reaction is known to be Ka = 5.64 once it reaches equilibrium. At equilibrium, 1.35 x 10-7 M of HCl and 1.2 M of Cl- can be found. What is the pH of the solution at equilibrium?
The following reaction occurs in water:
CH3OH <----> CH3O- + H+
Once the reaction was allowed to reach equilibrium, 0.5 M of CH3OH, 0.3 M of CH3O+, and 0.3 M of H+ were found in solution. What is the Ka of this solution?
The equilibrium constant for the auto-ionization of water at 10 °C is 3.0 x 10-15. What is the concentration of H3O+ ions in pure water at this temperature?
What is the pH of a 1.1M solution of nitrous acid (HNO2). The Ka of nitrous acid is 7.2 x 10-4
Ammonia(NH3) is a weak base, Kb = 1.8 x 10-5 and benzoic acid (C6H5CH2O-) is a weak acid, Ka = 6.3 x 10-5. A solution of ammonium benzoate (NH4)(C6H5CH2O-) is:
To 4.00 L of water was added 100 mg of HF (6.6 x 10-4). What is the pH?
CH3COOH is a weak acid with a Ka of 1.8 x 10-5. In what pH range would you most likely an aqueous solution of Na(CH3COO).
Acids and Bases: Henderson Hasselbalch
To a solution of acetic acid (Ka=1.8×10−5\rm K_a = 1.8 \times 10^{-5}) was added 0.1 mol of NaOH\rm NaOH. After the addition of NaOH\rm NaOH the volume was 1.4L and the pH\rm pH was 5.2.
a) This solution a buffer? (True/False)
b) Calculate the number of moles of acetic acid that were originally in solution. Report your answer in moles to three significant figures. Do not include units in the answer field
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
A solution of [CH3NH3][CH3COO] is made in an undergraduate laboratory. Use the information below to predict the acidity of the solution.
CH3COOH: Ka = 1.8 x 10-5
CH3NH3+: Ka = 2.3 x 10-11
Strength of Acids and Bases
What is the strongest acid below? What is the weakest acid?
Strongest acid:
Cl3CCOOH
Weakest acid:
HCN
Acids and Bases: Conjugate
Using the data in the table, which of the conjugate bases below is the strongest base?
AcidKaHOAc1.8×10−5HCHO21.8×10−4HClO3.0×10−8HF6.8×10−4\def\arraystretch{1.5} \begin{array}{c} \hline \rm Acid &&&K_a\\ \hline HOAc&&&1.8 \times 10^{-5}\\ HCHO_2&&&1.8 \times 10^{-4}\\ HClO &&& 3.0 \times 10^{-8}\\ HF &&& 6.8 \times 10^{-4}\\ \hline \end{array}
Acids and Bases: Acidity
Nitrous acid has a Ka of 7.1 x 10-5. What are [H3O+] and [NO2- ] in 0.50 M HNO2?
Which of the solutions below would have the HIGHEST pH value?
Methyl diethanolamine CH3N(C2H4OH)2 is a weak base which we will abbreviate as mdea. The conjugate acid, mdeaH+, has a pKa = 8.52.
Practice: Relating [H3O+], [OH-], pH and pOH
Compare two acids and their 1M solution:
Acid 1:Ka=1.0×10−3Acid 2:Ka=1.0×10−6\rm Acid\ 1:K_a=1.0\times10^{-3}\qquad Acid\ 2:K_a=1.0\times10^{-6}