Concept explainers
Determine the pH of (a) a
Interpretation:
The
Concept introduction:
The
measures the concentration of hydronium ions in a solution. The solution with high concentration of hydronium ions has a low
The
The ionization of the weak acid takes place as:
Answer to Problem 5QP
Solution:
a)
b)
Explanation of Solution
a)
Summarize the concentration at equilibrium as follows.
Consider
The equilibrium expression for a reaction is written as follows:
Here,
is the concentration of acetate ion,
is the concentration of acetic acid, and
Substitute the value of
value of
The value of
Concentration of
The
Substitute the value of
Hence, the
b) A solution that is
The acetate ions are formed from sodium acetate on dilution and from acetic acid by ionisation.
The equation for the ionisation of sodium acetate ion is as follows:
Here, acetate ion is formed on dilution in the solution. So, the concentration of the acetate ion from sodium acetate is 0.2 M and the sodium ion further does not take part in the reaction.
Summarize the concentration at equilibrium as follows:
Consider
The equilibrium expression for a reaction is written as:
Here,
Substitute the value of
value of
The value of
Concentration of
The
Substitute the value of
Hence, the
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Chapter 17 Solutions
Chemistry
- A 0.018 M solution of salicylic acid, HOC6H4CO2H, has the same pH as 0.0038 M HNO3solution. (a) Write an equation for the ionization of salicylic acid in aqueous solution. (Assume only the –CO2H portion of the molecule ionizes.) (b) What is the pH of solution containing 0.018 M salicylic acid? (c) Calculate the Ka of salicylic acid.arrow_forwardGiven that Ka’s for hydrofluoric acid (HF) and boric acid (H3BO3) are 6.3 × 10^–4 and 5.4 × 10^–10, respectively, calculate the pH of the following solutions: (a) The mixture from adding 50 mL 0.2 M HF to 50 mL 0.5 M sodium borate (NaH2BO3). (b) The mixture from adding an additional 150 mL 0.2 M HF to the solution in (a), i.e., a total of 200 mL 0.2 M HF was added to 50 mL 0.5 M NaH2BO3.arrow_forwardPropionic acid, HC3H5O2, has Ka= 1.34 x 10–5. (a) What is the molar concentration of H3O+ in 0.15 M HC3H5O2 and the pH of the solution? (b) What is the Kb value for the propionate ion, C3H5O2–? (c) Calculate the pH of 0.15 M solution of sodium propionate, NaC3H5O2. (d) Calculate the pH of solution that contains 0.12 M HC3H5O2 and 0.25 M NaC3H5O2.arrow_forward
- A mixture of 100 ml of 0.025 M NH3 and 150 mL of 0.025 M NH4Cl was made. For this problem you need to show your calculation for each number. (A) What is the resulting concentration of the two solutes after mixing the solutions?(B) What is the pH of the solution produced?(C) What will happen if additional 100 mL water is added to the mixture?arrow_forwardCalculate the pH change that results when 14 mL of 2.4 M HCl is added to 510. mL of each of the following solutions. (See the Acid-Base Table.) (a) pure water (b) 0.10 M CH3COO− (c) 0.10 M CH3COOH (d) a solution that is 0.10 M in each CH3COO− and CH3COOH.arrow_forward1.56 g of sodium acetate, NaCH;CO, has been 0.20 M ammonia, NH3, and 0.20 M ammonium 4. What is the pH of 0.15 M acetic acid to which 3. What is the pH of a solution that consists of aqueous solution of NH3? the same when you (c) add solid NaCl to a dilute aqueous solution of (b) add solid sodium acetate to a dilute a (a) add solid ammonium chloride to a dilute 1. Does the pH of the solution increase, decrease, or 17.1 and 17.2.) stay solution of acetic acid? aqueous NaOH? 2 Does the pH of the solution increase, decrease, or stay the same when you (a) add solid sodium oxalate, Na,C,O4, to 50.0 mL of 0.015 M oxalic acid, H,C,O4? (b) add solid ammonium chloride to 75 mL of 0.016 M HCl? (c) add 20.0 acetate, NaCH;CO2? of NaCl to 1.0 L of 0.10 M sodium *What is the pH of a solution that consists of chloride, NH4CI? added?arrow_forward
- Calculate the pH change that results when 15 mL of 2.0 M HCI is added to 580. mL of each of the following solutions (a) pure water 4.0-5.70 (b) 0.10 M CH3COO 4.04.28 (c) 0.10 M CH3COOH 4.0 (d) a solution that is 0.10 M in each CH3COO and CH3COOH. 4.0arrow_forward2.0 g of NaOH is dissolved in distilled water to prepare 100 mL solution. 20.0 mL of this solution reaches to the equivalence point when 25 mL of an acid solution containing 1.22 monoprotic weak acid is added. (a) Calculate the molar mass of the unknown acid. (b) After 15.0 mL of NaOH solution had been added during the titration, the pH was determined to be 4.7. What is the Ka of the unknown acid? (NaOH = 40.0 g/mol) unknownarrow_forwardA 0.1724-g sample of an unknown monoprotic acid was dissolved in 26.9 mL of water and titrated with 0.0623 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 19.8 mL. (a) Calculate the molar mass of the acid. (b) After 11.5 mL of base had been added during the titration, the pH was determined to be 5.66. What is the Ka of the unknown acid?arrow_forward
- Calculate the pH change that results when 15 mL of 2.0 M HCl is added to 560. mL of each of the following solutions. (See the Acid-Base Table attached.) (a) pure water (b) 0.10 M CH3COO− (c) 0.10 M CH3COOH (d) a solution that is 0.10 M in each CH3COO− and CH3COOH.arrow_forward(7) Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid( HC H O,,K=1.3x105) (b) 0.1000M sodium propanoate (Na C HỎ) (c) 0.1000M HC₂H₂O, and 0.1000M Nа С¸¸0₂ 3 5 52 (d) After 0.020 mol of HCl is added to 1.00 L solution of (a) and (b) above. (e) After 0.020 mol of NaOH is added to 1.00 L solution of (a) and (b) above.arrow_forwardCalculate the pH of the solution that results when 40.0 mL of 0.1250 M NH3 is (a) diluted to 20.0 mL with distilled water. (b) mixed with 20.0 mL of 0.250 M HCl solution. (c) mixed with 20.0 mL of 0.300 M HCl solution. (d) mixed with 20.0 mL of 0.200 M NH4C1 solution. (e) mixed with 20.0 mL of 0.100 M HC1 solution.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning