Titrations Chapter 13

Various Indicators Bromthymol Blue Phenolphthalein Methyl Red 6.2-7.6 Phenolphthalein 8.0-10.0 Methyl Red 4.4-6.2 Methyl Orange 3.1-4.4 Phenol Red 6.4-8.0

Neutralization Reactions In an aqueous solution, a reaction between an acid and a hydroxide base to produce salt and water. Example: (double replacement reaction) HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)

Titrations The controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of solution of unknown concentration. Procedure used to determine the unknown concentration of a solution. Indicators or conductivity probes can be used.

Types of titrations Strong acid and strong base Use bromthymol blue (6.2-7.6) Strong acid and weak base Use methyl orange (3.1-4.4) Weak acid and strong base Use phenolphthalein (8.0-10.0)

Equivalence Point During the titration, the point at which the concentration (in moles) of the standard solution is equal to the concentration (in moles) of the substance being titrated. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) Moles = moles

Endpoint The point at which the indicator changes color in the solution. This is very close to the equivalence point. Reaction is neutralized. HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)

Equivalence Point more base present pH E more acid present NaOH added (mL)

Steps to completing a titration See page 518 Steps 1-15. Results: Know the amount of acid added. (mL) Know the amount of base added. (mL) Know the concentration of standard solution (M) (moles/L) Know the balanced chemical equation. Can use these to figure out unknown concentration.

Molarity and Titration Determine the molarity of the NaOH solution. 20.0 mL of 0.100M HCl are titrated with also equal to 0.020 L of HCl and for every liter there are .100 mol HCl present. 19.5 mL of NaOH solution. Also 0.0195 L of solution.

Solving titrations ___HCl (aq) + ___NaOH (aq)  ___NaCl (aq) + ___H2O (l) Find the number of moles of acid used. (use molarity and volume) Find the number of moles of base needed. (use ratio and mol acid) __1_HCl (aq) + __1_NaOH (aq)  _1__NaCl (aq) + _1__H2O (l) 0.1 M x 0.020 L = 0.002 mol HCl used 0.002 mol HCl x 1 mol NaOH = 0.002 mol NaOH 1 mol HCl

Solving Titrations Find the concentration of based used. (use moles and liters). 0.002 mol NaOH = 0.103 M NaOH 0.0195 L