Concept explainers
(a)
Interpretation:
The plausible structures for
Concept introduction:
The main group elements tend to form oxyacid which are usually polyprotic acids like
(b)
Interpretation:
The pH at the first and second equivalence points needs to be determined assuming30.00 mL of 0.1240 M
Concept introduction:
The main group elements tend to form oxy-acids which are usually polyprotic acids like
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CHEMISTRY-TEXT
- For conjugate acidbase pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq) where Ka = 1.8 105 a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to part b. why do we classify the acetate ion (CH3CO2) as a weak base? Use an appropriate reaction to justify your answer. In general, as base strength increases, conjugate acid strength decreases. Explain why the conjugate acid of the weak base NH3 is a weak acid. To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is 1 106, calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids. To tie it all together, some instructors have students think of Li+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ as the conjugate acids of the strong bases LiOH, KOH. RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not technically correct, the conjugate acid strength of these cations is similar to the conjugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the correct response. The conjugate base of a weak acid is a_____base. The conjugate acid of a weak base is a_____acid. The conjugate base of a strong acid is a_____base. The conjugate acid of a strong base is a_____ acid. (Hint: Weak gives you weak and strong gives you worthless.)arrow_forwardEven though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)2?arrow_forwardWrite an equation for each of the following buffering actions. a. the response of a HPO42/PO43 buffer to the addition of OH ions b. the response of a HF/F buffer to the addition of OH ions c. the response of a HCN/CN buffer to the addition of H3O+ ions d. the response of a H3PO4/H2PO4 buffer to the addition of H3O+ ionsarrow_forward
- Phenol, C6H5OH, is a weak organic acid. Suppose 0.515 g of the compound is dissolved in enough water to make 125 mL of solution. The resulting solution is titrated with 0.123 M NaOH. C6H5OH(aq) + OH(aq) C6H5O(aq) + H2O() (a) What is the pH of the original solution of phenol? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH, and C6H5O? (c) What is the pH of the solution at the equivalence point?arrow_forwardAmmonium ion, NH4+, is a weak acid and its pKa is 9.25. (a) Write the balance chemical equation showing how ammonium ion acts like a weak acid. (b) What is the pH of a solution when initial concentration of ammonium ion is 0.020 M? (a) (b)arrow_forward8. (a) Calculate the percent ionization of a 0.20 M solution of the monoprotic acetylsalicylic acid (aspirin) for which Ka = 3.0 x 10-4 .(b)The pH of gastric juice in the stomach of a certain individual is 1.00. After a few aspirin tablets have been swallowed, the concentration of acetylsalicylic acid in the stomach is 0.20M. Calculate the % ionization of the acid under these conditions. What effect does the nonionized acid have on the membranes lining of the stomach? 9. Calculate the [H+], [C2O4-2], and [H2C2O4] in a 0.0010M H2C2O4 solution. (Ka = 5.4 x 10-2)arrow_forward
- Propionic 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_forwardCalculate the KHC8H4O4 titer of 0.125 M KOH. Give the chemical reactions involved.arrow_forward4) A highly toxic hydrogen cyanide (HCN) is a weak acid. A chemical engineer plans to determine pH of a 50 mL sample of HCN (0.10 M) in a titration process. To this end, she used 0.20 M NaOH as a titrant in varying volumes. Calculate the pH of the solution at the following points: (Ka for HCN=6.2×10-¹0) (a) Before addition of NaOH (initial pH), (b) After 10.00 mL of titrant addition, (c) After 25.00 mL of titrant addition, (d) After 50.00 mL of titrant addition.arrow_forward
- For each of the following cases, decide whether the pH is less than 7, equal to 7, or greater than 7. (a) Equal volumes of 0.10 M acetic acid, CH3CO2H, and 0.10 M KOH are mixed. (b) 25 mL of 0.015 M NH3 is mixed with 25 mL of 0.015 M HCl. (c) 150 mL of 0.20 M HNO3 is mixed with 75 mL of 0.40 M NaOH.arrow_forward(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C6H5COOH, Ka = 6.3 * 10-5). (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to20.0 mL of 0.100 M NH3.arrow_forwardConsider the titration of a weak acid that has a pKa = 4.00. Suppose a chemist was going to perform a titration on 50.0 mL of 0.050 0 M of the weak acid using 0.500 M NaOH. (A) What would be the pH of the solution after 0.00 mL of 0.500 M NaOH has been added?(B) What would be the pH halfway to the equivalence point of the titration?(C) What would be the pH at the equivalence point in the titration? (D) What would be the pH of the solution after 6.00 mL of 0.500 M NaOH has been added? Use the following reasoning when solving this problem. Because at the equivalence point the moles of strong base (0.500 M NaOH) added as a titrant equal the moles of weak acid, HA (50.0 ml of 0.050 M) in the solution, we can start by calculating the volume of 0.500 M NaOH needed to reach the equivalence point. To do this we simply equate the moles of HA to moles of NaOH by using the dilution formula. i.e. MaVa = MbVb where a represents HA and b represents NaOH. To calculate volume of NaOH needed to reach…arrow_forward
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