1. Acid Base Calculations Calculations involving pH Titration calculations

2. A 1.0 x 10 -4 M solution of HNO 3 has been prepared for a laboratory experiment. A) Calculate the [H 3 O + ] of this solution B) Calculate the [OH - ] HNO 3 (l) + H 2 O(l) H 3 O + (aq) + NO 3 - (aq) We can assume 100% ionization (strong acid) so HNO 3 concentration equals H 3 O + concentration [H 3 O + ][OH - ] = 1.0 x 10 -14 [H 3 O + ] [H 3 O + ] [OH - ] = 1.0 x 10 -14 [H 3 O + ] [OH - ] = 1.0 x 10 -14 = 1.0 x 10 -10 M 1.0 x 10 -4

3. What is the pH of a 1.0 x 10 -3 M NaOH solution? [H 3 O + ][OH - ] = 1.0 x 10 -14 [H 3 O + ] = 1.0 x 10 -14 [OH - ] [H 3 O + ] = 1.0 x 10 -14 = 1.0 x 10 -11 1.0 x 10 -3 pH = -log [H 3 O + ] = - log (1.0 x 10 -11 ) = 11.0

4. The pH of a solution is measured and determined to be 7.52. A) What is the hydronium ion concentration? B) What is the hydroxide ion concentration? C) Is the solution acidic or basic? [H 3 O + ][OH - ] = 1.0 x 10 -14 [OH - ] = 1.0 x 10 -14 [H 3 O + ] [OH - ] = 1.0 x 10 -14 = 3.3 x 10 -7 M 3.02 x 10 -8 pH = -log [H 3 O + ] log [H 3 O + ] = -pH [H 3 O + ] = antilog (-pH) = antilog (-7.52) = 3.02 x 10 -8 M

5. Titration Calculation In a titration, 27.4 ml of a 0.0154 M NaOH is added to 20.0 mL sample of HCl solution of unknown concentration. What is the molarity of the acid solution? HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) SA SB Neutralization First we need to figure moles of NaOH used to neutralize HCl in reaction. 0.0274 L x 0.0154 mol/L= 4.22 x 10 -4 mol of NaOH

6. Titration Calculation HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) = 4.22 x 10 -4 mol of NaOH = 4.22 x 10 -4 mol of HCl M of HCl = 4.22 x 10 -4 mol of HCl.0200 L M of HCl =.021 M

7. Titration Endpoint How can you tell when you have added enough NaOH?

8. Titration Curve