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pH Calculation and Acid-Base Analysis

Calculate the pH of a solution with a given [OH-] concentration and determine if the solution is acidic or basic. Learn about neutralization reactions and titrations.

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pH Calculation and Acid-Base Analysis

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  1. Do Now • a) What is the pH of a solution if the [OH-] = 4.0x10-11M? b) is this solution acidic or basic? ______________________

  2. Day 5 - Notes Unit: Acids, Bases, and Solutions Neutralization Reactions and Titrations

  3. After today you will be able to… • Explain what a neutralization reaction is • Write the corresponding formulas in neutralization reactions • Calculate an unknown concentration of an acid or base using a titration

  4. Neutralization Reactions A neutralization reaction is a double replacement reaction between an acid and a base to produce water and a salt (an ionic compound).

  5. In a titration, a solution of known concentration (called the standard solution) is used to determine the concentration of an unknown solution.

  6. Titrations A known quantity of one solution is measured out, and the other solution is added from the buret until the two solutions have neutralized each other. The point at which two solutions have neutralized each other is called the endpoint of the titration.

  7. Titrations An indicator is used to mark the endpoint of the titration. • Phenolphthalein:Is an indicator often used in acid-base titrations. It is colorless in acid and pink in base.

  8. Titration Setup

  9. Titration Calculations To calculate the unknown concentration of an acid or base using titrations, the following steps are used: • Write the balanced chemical equation. • Identify and label the numbers given by the problem • Re-write your formula to solve for the missing variable

  10. Example 0.037L of a 0.195M HBr solution is neutralized by 0.0536L of an unknown KOH solution. What is the concentration of the KOH? ___ HBr + ___ K(OH)  H(OH) 1 1 1 1 ___ H2O + ___ KBr MB=? MA=0.195M VB= 0.0536L VA=0.037L MA x VA =MB x VB 0.195M x MB 0.0536L x 0.037L = 0.0536L MB 0.13 M =

  11. Example 10.8 mL of an unknown HCl solution is neutralized by 0.0216L of a 0.950 M NaOH solution. What is the concentration of the HCl? ___ HCl + ___ Na(OH)  H(OH) 1 1 1 1 ___ H2O + ___ NaCl MB=0.950 M MA= ?M **Units are NOT the same. You have to make both mL or both L VB= 0.0216L VA=10.8 mL MA x VA =MB x VB MA x 0.95M x 0.0108L = 0.0216L 0.0108L MA = 1.9 M

  12. Questions?Begin HW5

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