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GENERAL CHEMISTRY-MOD.MASTERINGCHEM.
- 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_forwardAcrylic acid is used in the polymer industry in the production of acrylates. Its K, is 5.6 X 10“’. What is the pH of a 0.11 M solution of acrylic acid, CH2CHCOOH?arrow_forwardMost naturally occurring acids are weak acids. Lactic acid is one example. CH3CH(OH)CO2H(s)+H2O(l)H3O+(aq)+CH3CH(OH)CO2(aq) If you place some lactic acid in water, it will ionize to a small extent, and an equilibrium will be established. Suggest some experiments to prow that this is a weak acid and that the establishment of equilibrium is a reversible process.arrow_forward
- 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_forwardOne half liter (500. mL) of 2.50 M HCl is mixed with 250. mL of 3.75 M HCl. Assuming the total solution volume after mixing is 750. mL, what is the concentration of hydrochloric acid in the resulting solution? What is its pH?arrow_forwardThe hydrogen phthalate ion, C8HsO4, is a weak acid with Ka = 3.91 106. C8H5O4(aq)+H2O(l)C8H4O42(aq)+H3O+(aq) What is the pH of a 0.050 M solution of potassium hydrogen phthalate. KC8H5O4? Note: To find the pH for a solution of the anion, we must take into account that the ion is amphiprotic. It can be shown that, for most cases of amphiprotic ions, the H3O+ concentration is [H3O+]=Ka1Ka2 For phthalic acid, C8H6O4 is Ka1 is 1.12 103, and Ka2 is 3.91 106.arrow_forward
- 1. Calculate the pH of a 0.594 M aqueous solution of pyridinium chloride, C5H5NHCl. The Kb of pyridine is 1.8 x 10-9. The answer must be given with the correct number of significant figures. 2. How many grams of NaF must be added to 60.0 mL of 0.500 M HF(aq) to adjust the pH to 3.40? Assume no volume change. The pKa of HF is 3.14.arrow_forwardCalculate the pH of 0.10 M (COOH)2 (aq), oxalic acid. Ka1 = 5.9 × 10–2 ; Ka2 = 6.4 × 10–5.arrow_forwardSuppose that, instead of using NaOH, a base such as Ba(OH)2 had been used. What changes in the calculations would then have to be made to determine the molar concentrations of the base?arrow_forward
- Given 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_forwardCalculate the pH of the resulting solution if 26.0 mL of 0.260 M HCl(aq) is added to 36.0 mL of 0.260 M NaOH(aq). pH = Calculate the pH of the resulting solution if 26.0 mL of 0.260 M HCI(aq) is added to 16.0 mL of 0.360 M NaOH(aq). pH IIarrow_forwardWhat is the pH of a solution made by diluting 24 mL of 0.12 M HClO4(aq) to a total volume of 250 mL?arrow_forward
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