Download presentation

Presentation is loading. Please wait.

Published byShannon Tate Modified over 9 years ago

1
Acid/Base Indicators Substance that changes color in the presence of an acid or a base – Red or Blue Litmus – Phenolphthalein (phth) – Bromothymol blue (blue food coloring) – Red cabbage juice Episode 1102

2
Strong acids: – Dissociate completely in water – Ex: HCl Weak Acids: – Dissociate partially in water – Ex: HC 2 H 3 O 2 or vinegar Episode 1102

3
Strong bases: – Dissociate completely in water – Ex: NaOH Weak Bases: – Dissociate partially in water – Ex: NH 4 OH or ammonia solution Episode 1102

4
Acid/Base Concentration pH = - log [H+] 0 -----ACID-----7-----BASE-----14 Episode 1102

5
Determine the pH of a solution of HCl that has a molarity of 1 x 10 -4 M. HCl is a strong acid- know it completely dissociates. HCl H + + Cl - 1 mole HCl = 1 mol H + 1 x 10 -4 M HCl = 1 x10 -4 M H + pH = -log (1x 10 -4 ) pH = 4 Episode 1102

6
Calculate the pH for a solution of HNO 3 with a molarity of 1 x 10 -3 M. Strong acid – completely dissociates HNO 3 H + + NO 3 - 1 mol HNO 3 = 1 mol H + 1 x 10 -3 M HNO 3 = 1 x 10 -3 M H + pH = -log (1 x 10 -3 ) pH = 3 Episode 1102

7
Calculate the pH for a solution of H 2 SO 4 with a molarity of 1 x 10 -4 M. Strong acid – completely dissociates H 2 SO 4 2H + + SO 4 -2 1 mol H 2 SO 4 = 2 moles H + 1 x 10 -4 M H 2 SO 4 = 2(1 x 10 -4 M H + ) 2(1 x 10 -4 M H + ) = 2 x 10 -4 M H + pH = -log (2 x 10 -4 ) pH = 3.7 Episode 1102

8
Self Ionization of Water [H+][OH-] = 1 x 10 -14 Origin of pH scale – (-log[H+]) + (-log[OH-]) = -log (1 x 10 -14 ) – pH + pOH = 14 Episode 1102

9
Calculate the pH of a solution of NaOH with a molarity of 3.0 x 10 -2 M. Notice NaOH – acid or base? Strong bases – completely dissociates NaOH Na + + OH - 1 mol NaOH = 1 mol OH - 3.0 x 10 -2 M NaOH = 3.0 x 10 -2 M OH - [H + ][OH - ] = 1 x 10 -14 [H + ](3.0 x 10 -2 ) = 1 x 10 -14 [H + ] = 3.3 x 10 -13 pH = -log [H + ] pH = -log (3.3 x 10 -13 ) pH = 12.5 Episode 1102

10
Find the pH for a solution of Ca(OH) 2 with a molarity of 1 x 10 -4 M. Notice Ca(OH) 2 – acid or base? Strong bases – completely dissociates Ca(OH) 2 Ca +2 + 2OH - 1 mol Ca(OH) 2 = 2 mol OH - 1 x 10 -4 M Ca(OH) 2 = 2(1 x 10 -4 M OH - ) [OH-] = 2 x 10 -4 M OH - [H + ][OH - ] = 1 x 10 -14 [H + ](2 x 10 -4 ) = 1 x 10 -14 [H + ] = 5 x 10 -11 pH = -log [H + ] pH = -log (5 x 10 -11 ) pH = 10.3 Episode 1102

11
Calculate both the hydrogen ion concentration and the hydroxide concentration for an aqueous solution that has a pH of 4.0. pH = -log [H + ] 4.0 = -log [H + ] Log is a function of the number 10, so … -4.0 = log [H + ] 10 -4 = [H + ] or 1 x 10 -4 M [H + ] [H + ][OH - ] = 1 x 10 -14 (1 x 10 -4 )[OH - ] = 1 x 10 -14 [OH - ] = 1 x 10 -10 M Episode 1102

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google