Concept explainers
Interpretation:
Among the given graphs, the graph that best represents the relationship between pH of a acidic salt solution and the
Concept Information:
Base ionization constant
The equilibrium expression for the ionization of weak base
Where
pH and pOH:
The pOH scale is the reverse of pH scale
To Explain: The diagram that best represents the relationship between pH of a acidic salt solution and
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Chemistry: Atoms First
- 13. A 60.00 mL sample of 0.075 M sodium benzoate (NaC7H5O2) was titrated with 0.050 M HCl. What is the pH of the solutionafter 10.00 ml of HCl is added?(a) 4.19(b) 5.09(c) 5.74(d) 6.2414. What is the ratio of moles of benzoate (C7H5O2‒) to benzoic acid (HC7H5O2) in the solution that results from thecombination of the NaC7H5O2 and HCl in the problem above?(a) 8(b) 0.125(c) 0.0040(d) 0.00050arrow_forward(a) Calculate the acetate ion concentration in a solution prepared by dissolving 8.70×10-3 mol of HCl(g) in 1.00 L of 1.10 M aqueous acetic acid (Ka = 1.80×10-5). (b) Calculate the pH of the above solution. Give your answer to two decimal places.arrow_forwardB ONLY (a)How many ml of 0.5 M HCl must be added to 100.0 ml of a 0.2 M Imidazole (C3H4N2 ; Kb=6.3x10-8) solution to obtain a pH of 5.52? (b)Is this solution a buffer?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_forwardDetermine the pH of each of the following solutions.(a) 0.246 M hydrocyanic acid (HCN) (weak acid with Ka = 4.9e-10).(b) 0.228 M propionic acid (weak acid with Ka = 1.3e-05).(c) 0.850 M pyridine (weak base with Kb = 1.7e-09).arrow_forwardThe ion HTe− is an amphiprotic species; it can act as either an acid or a base.(a) What is Ka for the acid reaction of HTe− with H2O?(b) What is Kb for the reaction in which HTe− functions as a base in water?(c) Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe−] in a 0.10 M solution of H2Te.arrow_forward
- A solution is prepared at 25 °C that is initially 0.057M in dimethylamine ((CH3), NH), : a weak base with K,=5.4 × 10 and 0.29M in dimethylammonium bromide ((CH,), NH,Br). Calculate the pH of the solution. Round your answer to 2 decimal places. 2 pH = |||arrow_forwardDon’t ever mix acid with cyanide (CN) because it liberates poisonous HCN(g). But, just for fun, calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN with (a) 4.20 mL of 0.438 M HClO4. (b) 11.82 mL of 0.438 M HClO4. (c) What is the pH at the equivalence point with 0.438 M HClO4?arrow_forwardThe major component of vinegar is acetic acid, CH3COOH. Its Ka is 1.8 × 10-5 . One student used 1.000 M NaOH to titrate 25.00 mL vinegar. At the end point, 21.82 mL NaOH was used. (a) What is the concentration of CH3COOH in vinegar? (b) What is the pH of the solution at the end point? (c) What indicator(s) the student should use in this titration? Explainarrow_forward
- (a) Using the expression Ka=[H+][A−]/[HA], explain how to determine which solution has the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq). Do not perform any numerical calculations. (b) Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq) ? Justify your answer including the calculation of percent ionization for each solution.arrow_forwardA student is provided with a 0.1 M stock solution of NaOH. Student was then asked to mix 50.0 mL of this stock solution of NaOH with 450.0 mL of water to prepare a dilute solution of NaOH. (A) Calculate the molarity of the diluted solution. (B). Calculate the hydronium ion concentration, [H3O+] in the final diluted solution. (8 points) (C ) Calculate the pH of the diluted NaOH solution. (D) Is the final diluted solution acidic, basic or neutral?arrow_forwardA buffer is prepared by adding 4.8 g of (NH4)2SO4 to 425 mL of 0.258 M NH3. Assuming that the volume stays constant, what is pH of the buffer solution? Consider: Kb (NH3) = 1.8×10–5, and Molar Mass of (NH4)2SO4 = 132.14 g/mol. (A) 10.04 (B) 5.22 (C) 9.44 (D) 4.93 (E) 1.75arrow_forward
- Introductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage Learning