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Unit: Acids, Bases, and Solutions

Day 5 - Notes. Unit: Acids, Bases, and Solutions. Neutralization Reactions and Titrations. After today you will be able to…. Explain what a neutralization reaction is Write the corresponding formulas in neutralization reactions

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Unit: Acids, Bases, and Solutions

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

  2. 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

  3. Neutralization Reactions A neutralization reaction is a double replacement reaction between an acid and a base to produce water (NUETRAL pH )and a salt (an ionic compound that does not affect the pH).

  4. Neutralization Examples Predict the products in words and formulas, and balance! (Acid + Base ----- water + Salt ) Predict in formulas only and balance. __ Ca(OH)2 + __ H3(PO4)  hydrogen sulfite sulfurous acid potassium hydroxide potassium sulfite water + +  Ca+2 (PO4)-3 H(OH) 3 2 __ H2O 1 __ Ca3(PO4)2 6 + Ca 1231 3112 3 (OH) 6 6 H 6 6 (PO4) 2

  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. • Convert the known molarity into moles using the molarity formula: M=mol/L. • Convert moles of known into moles of unknown using the mole ratio. • Using the molarity equation, use moles of unknown and divide by volume to get molarity.

  10. Example 0.025L of a 0.28M HF solution is neutralized by 0.047L of an unknown Mg(OH)2 solution. What is the concentration of the Mg(OH)2? ___ HF + ___ Mg(OH)2 H(OH) 1 2 2 1 ___ H2O + ___ MgF2 0.025L 0.047L ? M 0.28M Step 1: Step 2: mol 1 mol Mg(OH)2 0.007mol HF 0.28M = 0.0035 mol Mg(OH)2 x = 0.025L 2 mol HF mol HF = 0.007mol Step 3: 0.0035 mol 0.074M Mg(OH)2 = 0.047L

  11. 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 0.037L 0.0536L ? M 0.195M Step 1: Step 2: mol 1 mol K(OH) 0.0072mol HBr 0.195M = 0.0072 mol K(OH) x = 0.037L 1 mol HBr mol HBr = 0.0072mol Step 3: 0.0072 mol 0.13M K(OH) = 0.0536L

  12. Questions?Begin WS5

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