1. Acid-Base Reactions Adding a base to an acid neutralizes the acid ’ s acidic properties. This reactions is called a neutralization reaction. The products of a neutralization reaction are water and salt. Acid + base  water + salt A salt is an ionic compound that is composed of the anion from the acid and the cation from the base.

2. HNO 3 (aq) + NaOH (aq)  NaNO 3 (aq) + H 2 O (l) Cation from base Anion from acid

3. The balanced chemical equation for this reaction shows that 1 mol of nitric acid reacts with 1 mol of sodium hydroxide. If equal molar quantities of reactants are used, then the result is a neutral (pH 7) aqueous solution. For most neutralization reactions, there are no visible signs that a reaction is occurring. One way is to use an acid-base indicator, which is a substance that changes color in acidic or basic solutions.

4. H + (indicator)  H + + indicator one coloranother color

5. Acid-Base Indicators A common method for determining pH of a solution is to use an indicator Indicators respond to changes in [H + ] (change colour) Universal Indicator Cabbage Juice Litmus Paper

6. Phenolphthalein Bromothymol blue

7. Finding Concentration Example 13.84 mL of hydrochloric acid just neutralizes 25.00 mL of a 0.1000 mol/L solution of sodium hydroxide. What is the concentration of the hydrochloric acid? Write the balanced equation Calculate the amount of sodium hydroxide added, based on volume and concentration of solution Determine the amount of hydrochloric acid needed to neutralize the sodium hydroxide Find the concentration based on the amount and volume of hydrochloric acid solution needed.

8. Balanced equation HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l) Calculate NaOH Amount NaOH added (in mol) = Concentration x volume = 0.1000 mol/L x 0.025 00 L = 2.500 x 10 -3 mol

9. Determine HCl HCl reacts with NaOH in a 1:1 ratio, so there must be 2.500 x 10 -3 mol HCl Concentration (mol/L) HCl = Amount (mol) volume (L) = 2.500 x 10 -3 mol 0.01384 L = 0.1806 mol/L

10. Finding Volume Example What volume of 0.250 mol/L sulfuric acid is needed to react completely with 37.2 mL of 0.650 mol/L potassium hydroxide? Write a balanced equation Calculate the amount (in mol) of potassium hydroxide Determine the amount of sulfuric acid needed Find the volume of sulfuric acid

11. Balanced equation H 2 SO 4(aq) + 2 KOH (aq)  K 2 SO 4 (aq) + 2 H 2 O (l) Calculate KOH Amount KOH added (in mol) = Concentration x volume = 0.650 mol/L x 0.0372 L = 2.41 x 10 -2 mol

12. Determine H 2 SO 4 1 H 2 SO 4 = x H 2 SO 4 2 KOH2.41 X 10 -2 1.21 x 10 -2 mol H 2 SO 4 Concentration (mol/L) HCl = Amount (mol) volume (L) Volume = Amount / Concentration = 1.21 x 10 -2 mol / 0.250 mol/L = 4.84 x 10 -2 L