What is the pH when Jeremiah has added 18.0 mL of 0.275 M NaOH to 20.0 mL of 0.500 M HCIO3? HCIO3 + NaOH → NaCIO3 + H₂O pH = [?] pH at 18.0 mL base Enter
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
![**Titration Calculation Example**
**Question:**
What is the pH when Jeremiah has added 18.0 mL of 0.275 M NaOH to 20.0 mL of 0.500 M HClO₃?
**Chemical Equation:**
HClO₃ + NaOH → NaClO₃ + H₂O
**Calculation:**
pH = [?]
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**Explanation:**
Let's break down the information provided and solve for the pH step-by-step:
1. Understand the reaction: An aqueous solution of strong acid HClO₃ is reacting with a strong base NaOH.
2. Calculate the moles:
- Moles of HClO₃ (Weak Acid): Molarity × Volume = 0.500 M × 20.0 mL/1000 mL = 0.0100 moles
- Moles of NaOH (Strong Base): Molarity × Volume = 0.275 M × 18.0 mL/1000 mL = 0.00495 moles
3. Determine the reaction's completion:
- Since we are adding fewer moles of NaOH than HClO₃, the base will be the limiting reagent, and the reaction will proceed until all NaOH is consumed.
- Moles of HClO₃ remaining after the reaction = Initial moles of HClO₃ - Moles of NaOH added = 0.0100 moles - 0.00495 moles = 0.00505 moles
4. Calculate the concentration of the remaining HClO₃ in the final mixture:
- Total volume of the mixture = 20.0 mL HClO₃ + 18.0 mL NaOH = 38.0 mL = 0.038 L
- Concentration of HClO₃ = Moles / Volume = 0.00505 moles / 0.038 L ≈ 0.13289 M
5. Finally, determine the pH. Since HClO₃ is a strong acid, pH ≈ -log[H⁺]
- [H⁺] = Concentration of HClO₃ = 0.13289 M
- pH = -log(0.13289) ≈](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8eb77797-9ac9-400b-9a5e-2bcfb6c77ad3%2F2b90886a-54a1-4d24-ad14-bff09cad6862%2Ft8238ib_processed.jpeg&w=3840&q=75)
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