12. Sodium Azide (NaN3) is sometimes added to water to kill bacteria. Calculate the pH ofa 0.010 mol/L solution of NaN 3. (6 Marks) [T] Ka=1.9x1056 points

Step 1: The chemical equation for the ionization of the azide ion in water isBased on the given…

Answered: 12. Sodium Azide (NaN3) is sometimes…

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter13: Acids And Bases

Section: Chapter Questions

Problem 76QAP: A solution of sodium cyanide, NaCN, has a pH of 12.10. How many grams of NaCN are in 425 mL of a...

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A buffer is prepared in which the ratio [ H2PO4 ]/[ HPO42 ]is 3.0. (a) What is the pH of this buffer? (b) Enough strong acid is added to convert 15% of HPO42- to H2PO4-. What is the pH of the resulting solution? (c) Enough strong base is added to make the pH 7.00. What is the ratio of [H2PO4-] to [HPO42-] at this point?
Acrylic 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?
You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the pH of the solution, and how would you use this information? How does the pH of the solution of just the HA compare with that of the final mixture? Explain.
Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation 2NaOCN(s)+H2C2O4(s)2HNCO(l)+Na2C2O4(s) Upon isolating pure HNCO(l), an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a l00.-mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and H2C2O4, assuming all of the HNCO produced is dissolved in solution? (Ka of HNCO = 1.2 l04.)
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.)
8-60 How is the buffer capacity affected by the ratio of the conjugate base to the conjugate acid?
A solution of sodium cyanide, NaCN, has a pH of 12.10. How many grams of NaCN are in 425 mL of a solution with the same pH?
. How is the strength of an acid related to the fact that a competition for protons exists in aqueous solution between water molecules and the anion of the acid?
Define a buffer solution. What makes up a buffer solution? How do buffers absorb added H+ or OH with little pH change? Is it necessary that the concentrations of the weak acid and the weak base in a buffered solution be equal? Explain. What is the pH of a buffer when the weak acid and conjugate base concentrations are equal? A buffer generally contains a weak acid and its weak conjugate base, or a weak base and its weak conjugate acid, in water. You can solve for the pH by setting up the equilibrium problem using the K.a reaction of the weak acid or the Kb reaction of the conjugate base. Both reactions give the same answer for the pH of the solution. Explain. A third method that can be used to solve for the pH of a buffet solution is the HendersonHasselbalch equation. What is the HendersonHasselbalch equation? What assumptions are made when using this equation?
Sketch the titration curve for a weak acid titrated by a strong base. When performing calculations concerning weak acidstrong base titrations, the general two-slep procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoichiometry part of the problem? What is assumed about this reaction? At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH? Why is pH 7.0 at the equivalence point of a weak acid-strong base titration? Does the pH at the halfway point to equivalence have to be less than 7.0? What does the pH at the halfway point equal? Compare and contrast the titration curves for a strong acidstrong base titration and a weak acidstrong base titration.
6. A buffer solution is formed by adding 0.500 mol of sodium acetate (CH3CO2NA) and 0.500 mol of acetic acid (CH3CO2H) to 1.00 L H,0. What is the pH of the solution at equilibrium? (K, = %3D 1.8 x 10°)
What is the pH of a 0.570 M solution of Ca(NO2)2 (Ka of HNO2 is 4.5 x 10-*)?

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12. Sodium Azide (NaN3) is sometimes added to water to kill bacteria. Calculate the pH of a 0.010 mol/L solution of NaN 3. (6 Marks) [T] Ka = 1.9x1056 points