Practice Questions:1) You have an unknown weak acid HA at a concentration of 5% (v/v), a density of 1.05 g/mL, and amolar mass of 60.a. Convert the %(v/v) to molarity.b. Fill in the following ICE Table:[HA][H*][A']InitialYour answer to laChangeEquilibrium0.00397c. Calculate HA's Ka. What acid is it?d. What is the %-ionization of HA?

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Answered: Practice Questions: 1) You have an…

Living by Chemistry

2nd Edition

ISBN:9781464142314

Author:Angelica M. Stacy

Publisher:Angelica M. Stacy

ChapterU6: Showtime: Reversible Reactions And Chemical Equilibrium

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A diprotic acid, H2B(MM=126g/moL), is determined to be a hydrate, H2B xH2O. A 10.00-g sample of this hydrate is dissolved in enough water to make 150.0 mL of solution. Twenty-five milliliters of this solution requires 48.5 mL of 0.425 M NaOH to reach the equivalence point. What is x?
Two students were asked to determine the Kb of an unknown base. They were given a bottle with a solution in it. The bottle was labeled aqueous solution of a monoprotic strong acid. They were also given a pH meter, a buret, and an appropriate indicator. They reported the following data: volume of acid required for neutralization = 21.0 mL pH after 7.00 mL of strong acid added = 8.95 Use the students' data to determme the Kb of the unknown base.
Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
Enough water is added to the buffer in Question 30 to make the total volume 5.00 L. (a) Calculate the pH of the buffer. (b) Calculate the pH of the buffer after adding 0.0250 mol of HCl to 0.376 L of the buffer. (c) Calculate the pH of the buffer after adding 0.0250 mol of KOH to 0.376 L of the buffer. (d) Compare your answers to Question 30 (a-c) with your answers to (a-c) of this problem. (e) Comment on the effect of dilution on the pH of a buffer and on its buffer capacity. Â
A friend asks the following: Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH to form A. Thus the amount of acid (HA) is decreased, and the amount of base (A) is increased. Analogously, adding HCI to the buffered solution forms more of the acid (HA) by reacting with the base (A). Thus how can we claim that a buffered solution resists changes in the pH of the solution? How would you explain buffering to this friend?
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a. What is the percent ionization in 0.80 M NH3 (Kb Percent ionization = % % = % 1.8 × 10-5)? b. What is the percent ionization in 0.090 M NH3 (Kb Percent ionization = c. What is the percent ionization in 1.0 M CH3NH₂ (Kb = 4.38 × 10¯4)? Percent ionization = = 1.8 × 10-5)?
15:47 Tue Mar 7 1 4 Question 6 of 10 What is the molarity of ions in a 0.824 M solution of Li₂SO4 assuming the compound dissociates completely? 2 5 8 7 +/- Tap here or pull up for additional resources 1.684 M 3 6 9 0 ome x160 X C Submit x 100 < 1. m a. a 3. M 4. M 5. M D -W the 0 a. N 8.1 N 14 0 w ob ww -tra - 2₁ Sul 7. 2. % W ob Fra
(a) Calculate the percent ionization of 0.00280 M carbonic acid (Ka = 4.3e-07).% ionization = ______ %(b) Calculate the percent ionization of 0.00280 M carbonic acid in a solution containing 0.0200 M sodium hydrogen carbonate.% ionization =_______ %
7:50 A14 (4th quarter) pH and po... 3.24 x 10-3 1.09 x 10-10 M 4.325 1 x 10-4 M Part B. 1. The hydrogen ion concentration in a solution is measured to be 7.84 X 10-5 M. Calculate the following: a) pH b) pOH c) [OH-], M 2. You have 500.0 mL (convert to liter by dividing by 1,000) of a 4.5 x 10-¹¹M of HNO₂ solution. a. How many grams of HNO3 are present in this solution? Step 1: Using the formula of molarity, solve for the moles of solute: Moles solute = M x L Step 2: Solve the MM of HNO3 using the periodic table of elements: Step 3: Multiply Steps 1 and 2 results to get the grams of HNO3. b. Complete the chart using the given value of the concentration of [H*]. [OH-] [H+] 4.5 x 10-¹¹M pH POH 11.76
Canvas Question 52 EU A student attempt to execute the reaction below between H3CCN and KOH. -H pka = 25 Но Is KOH a sufficiently strong base for this process? O Yes. K = 2.00 x 10⁹ O Yes. Ke 5.0 x 10¹0 O No. K = 2.00 x 10-41 O No. K = 5.0 x 10-10 nation 53 What is its molar solubility? EC? H₂0 pka = 15.7 Pro

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