(2)When ammonium sulfate dissolves, both the anion and cation have acid-base reaction:(NH4)2SO4(s)< 2NH4* + SO4²-NH4* + NH3(aq) + H*SO42- + H2O → HSO4 + OH-Ksp = 276Ka = 5.70 ×10-10Кь 3 9.80 х 10-13(a)(b)(c)(Show your Calculation)Write a charge balance for this system.Write a mass balance for this system.Find the concentration of NH3(aq) if the pH is fixed at 9.25.

Questions are 1. Charge balance 2. Mass balance 3. NH3 concentration calculation

When ammonium sulfate dissolves, both the anion and cation have acid-base reaction: (NH4)2SO4(s)→ 2NH4* + SO4²- NH4* + NH3(aq) + H* SO,2- + H2O → HSO4 + OH- Ksp = 276 Ka = 5.70 ×10-10 Kb = 9.80 × 10-13 (а) (b) (c) (Show your Calculation) Write a charge balance for this system. Write a mass balance for this system. Find the concentration of NH3(aq) if the pH is fixed at 9.25.

When ammonium sulfate dissolves, both the anion and cation have acid-base reaction: (NH4)2SO4(s)→ 2NH4* + SO4²- NH4* + NH3(aq) + H* SO,2- + H2O → HSO4 + OH- Ksp = 276 Ka = 5.70 ×10-10 Kb = 9.80 × 10-13 (а) (b) (c) (Show your Calculation) Write a charge balance for this system. Write a mass balance for this system. Find the concentration of NH3(aq) if the pH is fixed at 9.25.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Calculate the pH and the equilibrium concentrations of HSeO3 and Se032 in a 0.0298 M selenious acid solution, H₂Se03 (aq). For H₂Se03, Kal = 2.4×10-³ and Kaz pH = [HSe03] = [Se03²¯] = M M 4.8x10-9 A
The pH scale depends on the value of Kw. What would the pH of neutral water be if the Kw was 5.54 x10-15? IMPORTANT: When entering your answer: •Enter the number only (no units) • Do not leave any spaces Use a leading zero before the decimal when necessary Report your number to 3 decimal places (regardless of the significant figures) • Correctly round your answer to the 3rd decimal place (allowed margin of error is only ± 0.001, so always use un- rounded numbers in your calculations) Examples: 0.123 or 9.876
Identify the acid-base reaction: Group of answer choices ClO (g) + ClO (g) → Cl2O2 (g) C2H4 (g) + HCl (g) → C2H5Cl (g) 2HCl (aq) + Na2CO3 (aq) → 2NaCl (aq) + CO2(g) + H2O (ℓ) I2 (aq) + 2S2O32- (aq) → 2I- (aq) + S4O62- (aq)
Calculate the concentration of each species present in a 0.025 M solution of phthalic acid, C6H4(CO2H)2. C6HĄ(CO2H),(aq) + H,O(1) = H;0*(aq) +C6H4(CO2H)(CO2)¯(aq) Ką=1.1×10-3 CoH4(CO,H)(CO2)-(aq)+ HzO(1)= H30*(aq) + CgHq(CO2),²-(aq) Ką=3.9x 10-6
0.75 mg/L of hypochlorous acid (HOCl) is added to water at 25 C. The pH of the water is adjusted to 6.54. What is the molar ratio of hypochlorite ion (OCl-) to hypochlorous acid? Show your work. Given: HOCL ↔ H+ + OCl- ; Ka = 10-7.54; MW’s: H = 1; O = 16; Cl = 35.45
In a 0.09 M solution of chlorous acid, what percentage of the acid is ionized to the chlorous anion, ClO2-? Ka = 1.1 x 10-2 include 3 significant figures in your answer
Phthalic acid (H 2 C 8 H 4 O 4 , abbreviated H 2 Ph ) is a diproticacid. Its ionization in water at 25°C takes place in twosteps: H 2 Ph ( a q ) + H 20 (1)2H 30+ ( a q ) + HPh – ( a q) Ка1-1.26 х 10- 3Н2 Ph - (аq)+H20(1) нзо+(аq) + Ph 2 - (aq) If 0.0100 mol of phthalic acid is dissolved per liter ofwater, calculate the equilibriumconcentrations of H 2 Ph , H P h-, Ph Ка 2-3.10х 10 - 6 2 -, and H 30 +.
Calculate the pH of a 0.051 M (C₂H5)2NH solution (Kb = 1.3 x 10-3). pH =
H2SO4(aq) + H2O(l) → H3O+ (aq) + HSO4 (aq) O H2O is the acid (proton acceptor); H2SO4 is the base (proton donor). OH2SO4 is the acid (proton acceptor); H2O is the base (proton donor). OH2SO4 is the acid (proton donor); H2O is the base (proton acceptor). O H2O is the acid (proton donor); H2SO4 is the base (proton acceptor).
Please provide the correct reagents labelled 1-10: