# Example Problem: Determining pH in a Titration## ObjectiveDetermine the pH at the point in the titration of 40.0 mL of 0.200 M hydrazine solution (H₂NNH₂) with 0.100 M nitric acid (HNO₃) after 80.0 mL of the strong acid has been added. The Kb value for H₂NNH₂ is 3.0 × 10⁻⁶.## ProcedureBased on the result of the acid-base reaction, set up the ICE (Initial, Change, Equilibrium) table in order to determine the unknown concentrations.### Reaction Equation\[ \text{H}_2\text{NNH}_3^+(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{H}_2\text{NNH}_2(aq) \]## ICE Table Format| \[ \] | \[\text{H}_2\text{NNH}_3^+(aq)\] | \[\text{H}_2\text{O}(l)\] | \[\text{H}_3\text{O}^+(aq)\] | \[\text{H}_2\text{NNH}_2(aq)\] ||-------|-----------------|---------------|-----------------|---------------|| **Initial (M)** | \[ \] | \[ \] | \[ \] | \[ \] || **Change (M)** | \[ \] | \[ \] | \[ \] | \[ \] || **Equilibrium (M)** | \[ \] | \[ \] | \[ \] | \[ \] |## Graphical ControlsAt the bottom, select from the pre-encoded concentration fields:- Initial Concentrations: - 0.100 - 0.200 - 0.0333 - 0.0667- Changes in Concentration: - \( +x \) - \( -x \)Final Equilibrium Concentrations:- \[ 0.100 + x \]- \[ 0.100 - x \]- \[ 0.200 + **Titration Question: Determining pH****Problem Statement:**Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 80.0 mL of the strong acid has been added. The value of Kb for H₂NNH₂ is 3.0 x 10⁻⁶.**Instructions:**Use the table below to determine the moles of reactant and product after the reaction of the acid and base.**Reaction Table:**| | H₂NNH₂(aq) | + | H⁺(aq) | → | H₂NNH₃⁺(aq) ||-------------------------------|-----------|---|--------|---|-------------|| **Before (mol)** | | | | | || **Change (mol)** | | | | | || **After (mol)** | | | | | |**Interactive Options:**- Below the table, there are several options with different values and expressions to select from, such as: - 0 - 0.100 - +x - -x - Numerical values like 1.00 x 10⁻³, 2.00 x 10⁻³, etc. **Note:**Make sure to use the given values alongside the expressions and equations to solve for the unknowns in the reaction table. Adjustments can be made using the "RESET" button if necessary.

given values : Kb =3×10-6pKb=-log (3×10-6) =5.52 for weak base hydrazine : molarity = 0.2 M…

Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH2 with 0.100 M HNO3 after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 106. 1 3 4 NEXT Use the table below to determine the moles of reactant and product ater the reaction the acid and base. H2NNH2(aq) H*(aq) H,NNH; *(aq) + Before (mol) Change (mol) After (mol)

Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH2 with 0.100 M HNO3 after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 106. 1 3 4 NEXT Use the table below to determine the moles of reactant and product ater the reaction the acid and base. H2NNH2(aq) H(aq) H,NNH; (aq) + Before (mol) Change (mol) After (mol)

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...

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