Part A Calculate the concentration of an aqueous solution of NaOH that has a pH of 10.87. Express your answer in molarity to two significant figures. —| ΑΣΦ [NaOH] = Submit Provide Feedback Request Answer ? Review | Const M

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Acid-Base Equilibria Assignment Example:**

**Problem Overview:**

This exercise focuses on calculating the concentration of an aqueous solution of sodium hydroxide (NaOH) based on its given pH level. The target pH is 10.87.

**Task:**

Calculate the concentration of NaOH in molarity (M), ensuring that your final answer has two significant figures.

**Input Area:**

- Text box to input the concentration value of NaOH.

**Buttons:**

- Submit: To input your answer.
- Request Answer: To seek help or answers if needed.

**Instructions:**

1. Understand the relationship between pH, pOH, and hydroxide ion concentration.
2. Use the formula: \[ \text{pOH} = 14 - \text{pH} \]
3. Convert pOH to hydroxide concentration using: \[ [\text{OH}^-] = 10^{-\text{pOH}} \]
4. Input the calculated concentration of NaOH in the box provided and submit.

**Feedback:**

- A "Provide Feedback" option is available to leave any comments or suggestions.

This exercise is a practical example of applying acid-base equilibrium concepts to determine the concentration of a basic solution given its pH value.
Transcribed Image Text:**Acid-Base Equilibria Assignment Example:** **Problem Overview:** This exercise focuses on calculating the concentration of an aqueous solution of sodium hydroxide (NaOH) based on its given pH level. The target pH is 10.87. **Task:** Calculate the concentration of NaOH in molarity (M), ensuring that your final answer has two significant figures. **Input Area:** - Text box to input the concentration value of NaOH. **Buttons:** - Submit: To input your answer. - Request Answer: To seek help or answers if needed. **Instructions:** 1. Understand the relationship between pH, pOH, and hydroxide ion concentration. 2. Use the formula: \[ \text{pOH} = 14 - \text{pH} \] 3. Convert pOH to hydroxide concentration using: \[ [\text{OH}^-] = 10^{-\text{pOH}} \] 4. Input the calculated concentration of NaOH in the box provided and submit. **Feedback:** - A "Provide Feedback" option is available to leave any comments or suggestions. This exercise is a practical example of applying acid-base equilibrium concepts to determine the concentration of a basic solution given its pH value.
**Part D**

**Problem Statement:**
Give the conjugate acid of the following Brønsted-Lowry base: \( \text{H}_2\text{AsO}_4^- \).

**Instruction:**
Express your answer as a chemical formula.

**Answer Input Field:**
There is an input box where the answer should be typed in a chemical formula format. Below the input box, there is an option: "A chemical reaction does not occur for this question."

**Submission:**
There are buttons labeled "Submit" and "Request Answer" for submitting the answer or asking for help.

---

**Explanation for Educational Purpose:**

In Brønsted-Lowry theory, acids are proton donors and bases are proton acceptors. The conjugate acid is formed by adding a proton (\( \text{H}^+ \)) to the base. For the given base \( \text{H}_2\text{AsO}_4^- \), its conjugate acid is \( \text{H}_3\text{AsO}_4 \).
Transcribed Image Text:**Part D** **Problem Statement:** Give the conjugate acid of the following Brønsted-Lowry base: \( \text{H}_2\text{AsO}_4^- \). **Instruction:** Express your answer as a chemical formula. **Answer Input Field:** There is an input box where the answer should be typed in a chemical formula format. Below the input box, there is an option: "A chemical reaction does not occur for this question." **Submission:** There are buttons labeled "Submit" and "Request Answer" for submitting the answer or asking for help. --- **Explanation for Educational Purpose:** In Brønsted-Lowry theory, acids are proton donors and bases are proton acceptors. The conjugate acid is formed by adding a proton (\( \text{H}^+ \)) to the base. For the given base \( \text{H}_2\text{AsO}_4^- \), its conjugate acid is \( \text{H}_3\text{AsO}_4 \).
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