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
- Calculate the pH change that results when 11 mL of 5.1 M NaOH is added to 790. mL of each the following solutions. Use the Acid-Base Table. (a) pure water 4.0 5.84 (b) 0.10 M NH4CI 4.0 5.43 (c) 0.10 M NH3 4.0✔ X (d) a solution that is 0.10 M in each NH4+ and NH3 4.0✔arrow_forwardA biology experiment requires the use of a nutrient fluid buffered at a pH of 4.85, and 659 mL of the solution is needed. It has to be buffered to be able to hold the pH to within =0.10 pH units of 4.85 even if 5.00 x 10 mol of OH or 5.00 x 10 mol of H* ion enter. Record all your answers to three (3) significant figures. Pick the best acid and its sodium salt that could be used to prepare the solution. O Hydrocyanic acid and sodium cyanide (pK, = 9.21). O Nitrous acid and sodium nitrite (pk, = 3.15). Acetic acid and sodium acetate (pK, = 4.74). O Hydrofluoric acid and sodium fluoride (pK, = 3.17). Hint Calculate the minimum number of grams of the acid and its sodium salt that are needed to prepare the buffer solution. Mass acid i Mass sodium salt = i Hint Calculate the molar concentration of the acid and its sodium salt that are needed to prepare the buffer solution. Concentration acid = M Concentration sodium salt = M Hint onarrow_forward13. 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_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
- The 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_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_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_forward
- If the pH of a 0.200 M solution of the CICH2O¯ ion is 12.05, then what is the value for the basicity constant, Kp, of CICH20-? You do not need to make a simplifying assumption CICH,0-(aq) + H2O(1) СICH-OОH(aq) + ОН (аq) K½ = ?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_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