### Solving the pH of a 0.10 M Aqueous Solution of a Weak Base BTo determine the pH of a 0.10 M aqueous solution of the weak base B, follow these steps and fill in the blanks in the table and equilibrium expression without ignoring \( x \).#### Equilibrium Expression- **B(aq)**: The concentration of the weak base in aqueous solution.- **H₂O(l)**: Water, the solvent, playing its usual role in dissociation.- **HB⁺(aq)**: The conjugate acid of the weak base after gaining a proton.- **OH⁻(aq)**: Hydroxide ion, indicating the basic nature of the solution.### Table of Concentrations (M)Below is the general table format to be filled out, showing the relationships between initial concentrations, changes, and equilibrium concentrations.| | B(aq) | + | H₂O(l) | ⇌ | HB⁺(aq) | + | OH⁻(aq) ||---------|------------|---|--------|---|---------|---|----------|| **Initial** | [Blank 1] | | | | [Blank 2] | | [Blank 3] || **Change** | [Blank 4] | | | | [Blank 5] | | [Blank 6] || **Equilibrium** | [Blank 7] | | | | [Blank 8] | | [Blank 9] |### Equilibrium Constant Expression\[ K_b = \frac{[HB^+][OH^-]}{[B]} \]\[ pH = - \log [ x ] \]### Explanation of the Blanks - Fill in the blank values using the principles of chemical equilibrium where initial concentrations decrease or increase by a factor denoted as \( x \).### Instructions for Filling the Table1. **Identify initial concentrations:** - Initial concentration of \( B \): 0.10 M (Blank 1). - \( H₂O \) is not normally included in the expression as its concentration is large and constant (thus not provided). - Initial concentrations of \( HB^+ \) and \( OH^- \) are zero before dissociation starts (Blank 2 and Blank 3).2.

Answered: Image /qna-images/answer/272d0bf3-7218-4464-81af-9ff6e721aac6.jpg

Fill in the blanks to show how you would solve for the pH of a 0.10 M aqueous solution of the weak base B. For the table and the equilibrium expression, do not ignore x. 0.10 POH Initial Concentration (M) Change 0.10-x Equilibrium 2x² 0.10 - 2x 4x² B(aq) X (0.10 - x)² + 2x (0.10 - 2x)² H₂O(l) H + 0 4x HB*(aq) x² 0.10- x² + pH 0.10 + x OH(aq)

Fill in the blanks to show how you would solve for the pH of a 0.10 M aqueous solution of the weak base B. For the table and the equilibrium expression, do not ignore x. 0.10 POH Initial Concentration (M) Change 0.10-x Equilibrium 2x² 0.10 - 2x 4x² B(aq) X (0.10 - x)² + 2x (0.10 - 2x)² H₂O(l) H + 0 4x HB*(aq) x² 0.10- x² + pH 0.10 + x OH(aq)

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

A 2.00 L solution contains 0.545 M NH3 and 0.641 M NH4A. (The Kb of NH3 is 1.86⋅10−51.86⋅10-5). What is the pH of this solution? What is the pH of this solution after 0.121 mol of HCl are added to the solution? What is the pH of this solution after 0.242 mol of HCl are added to the solution?
Classify the following solutions as one of these types AND calculate the pH: strong acid, strong base, weak acid, weak base, buffer, neutral salt. a) 0.040 M NaF (Ka for HF = 7.2 x 10-4) b) 10 M HClO4 c) A solution made by mixing 10.0 mL of 0.072 M NaOH with 15.0 mL of 0.100 M acetic acid (pKa for acetic acid = 4.75) d) 0.453 M NaCl
A solution is prepared that is initially 0.45M in pyridine (CÂHÂN), a weak base, and 0.50M in pyridinium bromide (C5H₁NHBr). Complete the reaction table below, so that you could use it to calculate the pH of this solution. Use x to stand for the unknown change in [où¯]. You can leave out the symbol for molarity. initial change final [C,H,N] [C₂H₂NH*] [OH-] 0 0 0 0 DO X Ar
Hydrofluoric acid has a Ka of 6.6 x 10-4. 40.00 mL of 0.150 M sodium fluoride, NaF, is titrated against 0.125 M HBr. Give the pH at these four points: initial, 32.00 mL of acid have been added, at the equivalence point, and when 60.00 mL have been added.
Determine the pH of a buffer solution by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. Complete Parts 1-3 before submitting your answer. 3 > A solution is prepared with 0.55 M HNO2 and 0.75 M KNO2. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) -2x 0.75 -2x 0 1 HNO₂(aq) 0.55 + x 0.55 -X 0.55-x 0.55 0.55 - x 0.75 2 H₂O(l) III 0.55 + 2x 6.8 x 104 0.55 - 2x 1L +x H3O+(aq) 0.75 + x +x -X 0.75-x + NO₂ (aq) RESET +2x NEXT 0.75 + 2x +x
Determine the molar concentrations of Ca2+ and CI- in a 0.38 M calcium chloride solution. 1) 0.38 M Ca2+ and 0.38 M CI- O 2) 00.19 M Ca2+ and 0.38 M CI- 3) 0.38 M Ca2+ and 0.76 M CI- 4) 0.19 M Ca2+ and 0.76 M CI- O 5) None of the above
An analytical chemist is titrating 231.8 mL of a 0.9900M solution of acetic acid (HCH, CO,) with a 1.200M solution of NaOH. The p K, of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 131.2 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. alo pH %3| Ar Explanation Check © 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center | Accessibility CA
A chemist titrates 100.0 mL of a 0.5181 M carbonic acid (H,CO, solution with 0.7276 M NaOH solution at 25 °C. Calculate the pH at equivalence. The p K, of carbonic acid is 3.60. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.
Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer solution with an optimum pH of 3.2–3.5 a - NaNO 2 / HNO 2 (K a = 4.5 × 10 –4) b - NaOCl / HOCl (K a = 3.2 × 10 –8) c - NaCl / HCl d - NH 3 / NH 4Cl (K a = 5.6 × 10 –10) e - CH 3COO 2Na / CH 3COOH (K a = 1.8 × 10 –5)
What is the pH of a buffer solution that is 0.21 M methylamine and 0.20 M methylammonium chloride? K, (methylamine) = 4.4 × 10-4
Calculate the pH for each of the cases in the titration of 35.0 mL of 0.240 M KOH(aq) with 0.240 M HCl(aq). Note: Enter your answers with two decimal places. before addition of any HCl: after addition of 13.5 mL HCl: after addition of 21.5 mL HCl: after addition of 35.0 mL HCl: after addition of 41.5 mL HCl: after addition of 50.0 mL HCl:
A solution contains 0.351 M HA (Ka = 1.13⋅10−51.13⋅10-5) and 0.744 M NaA. What is the pH of this solution? What is the pH of this solution after 0.186 mol of HCl are added to 1.00 L of this solution? What is the pH of this solution after 0.372 mol of HCl are added to 1.00 L of this solution?