Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH2 with 0.100 M HNO; after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. 1 2 4 Based on the result of the acid-base reaction, set the ICE table in order to determine the dn unknown. H2NNH3*(at H20(1) =H;0*(aq) H2NNH2(aq Initial (M) Change (M) Equilibrium (M)
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH2 with 0.100 M HNO; after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. 1 2 4 Based on the result of the acid-base reaction, set the ICE table in order to determine the dn unknown. H2NNH3*(at H20(1) =H;0*(aq) H2NNH2(aq Initial (M) Change (M) Equilibrium (M)
Chemistry
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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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![**Question 16 of 20**
**Determine the pH at the point in the titration of 40.0 mL of 0.200 M \( \text{H}_2\text{NNH}_2 \) with 0.100 M \( \text{HNO}_3 \) after 80.0 mL of the strong acid has been added. The value of \( K_b \) for \( \text{H}_2\text{NNH}_2 \) is \( 3.0 \times 10^{-6} \).**
**Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown.**
**ICE Table:**
\[
\begin{array}{|c|c|c|c|c|}
\hline
& \text{H}_2\text{NNH}_3^+ (\text{aq}) & \text{H}_2\text{O (l)} & \rightleftharpoons & \text{H}_3\text{O}^+ (\text{aq}) \\ + & \text{H}_2\text{NNH}_2 (\text{aq}) \\
\hline
\text{Initial (M)} & & & & \\
\hline
\text{Change (M)} & & & & \\
\hline
\text{Equilibrium (M)} & & & & \\
\hline
\end{array}
\]
**Calculations:**
- Initial concentrations:
- \( \text{H}_2\text{NNH}_3^+ \): 0.0333 M
- \( \text{H}_2\text{NNH}_2 \): 0.0667 M
- Change in concentrations represented by \( \pm x \):
- \( \text{H}_3\text{O}^+ \): \( +x \)
- \( \text{H}_2\text{NNH}_3^+ \): 0.100 + x
- \( \text{H}_2\text{NNH}_2 \): 0.200 - x
- Equilibrium concentrations:
- \( \text{H}_2\text{NNH}_3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b31c76d-31c3-4f02-9c7e-c3848d4eba7b%2Fcc7433b5-91f0-459a-a010-d3e09ea0a160%2Fp1n1lmw_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 16 of 20**
**Determine the pH at the point in the titration of 40.0 mL of 0.200 M \( \text{H}_2\text{NNH}_2 \) with 0.100 M \( \text{HNO}_3 \) after 80.0 mL of the strong acid has been added. The value of \( K_b \) for \( \text{H}_2\text{NNH}_2 \) is \( 3.0 \times 10^{-6} \).**
**Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown.**
**ICE Table:**
\[
\begin{array}{|c|c|c|c|c|}
\hline
& \text{H}_2\text{NNH}_3^+ (\text{aq}) & \text{H}_2\text{O (l)} & \rightleftharpoons & \text{H}_3\text{O}^+ (\text{aq}) \\ + & \text{H}_2\text{NNH}_2 (\text{aq}) \\
\hline
\text{Initial (M)} & & & & \\
\hline
\text{Change (M)} & & & & \\
\hline
\text{Equilibrium (M)} & & & & \\
\hline
\end{array}
\]
**Calculations:**
- Initial concentrations:
- \( \text{H}_2\text{NNH}_3^+ \): 0.0333 M
- \( \text{H}_2\text{NNH}_2 \): 0.0667 M
- Change in concentrations represented by \( \pm x \):
- \( \text{H}_3\text{O}^+ \): \( +x \)
- \( \text{H}_2\text{NNH}_3^+ \): 0.100 + x
- \( \text{H}_2\text{NNH}_2 \): 0.200 - x
- Equilibrium concentrations:
- \( \text{H}_2\text{NNH}_3
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