Chapter 19 DHS Chemistry

There are many theories and definitions of acids and bases.

An acid is a molecule or ion that acts as a proton (hydrogen ion) donor. HCl + H20  H30+ + Cl- acid

A base is a molecule or ion that acts as a proton (hydrogen ion) acceptor. HCl + NH3  NH4+ + Cl- base

An acid produces hydronium (H30+) ions when dissolved in water. Ex. HCl + H2O  Cl- + H3O+

A base produces hydroxide (OH- ) ions when dissolved in water. Ex. NH3 + H2O  NH4+ + OH-

Acids that can contain multiple hydrogens to donate. Match the terms with the probable acid monoprotic H2SO4 diprotic H3PO4 polyprotic HCl

Acids that contain 1 ionizable hydrogen are monoprotic acids. Ex. HCl, HBr

Acids that contain 2 ionizable hydrogens are diprotic acids. Ex. H2SO4

Acids that contain 3 ionizable hydrogens are triprotic/polyprotic acids. Ex. H3PO4

Some substances can act as both an acid and a base – these substances are considered amphoteric. EX: HCl + H20  H30+ + Cl-  water as a base acid base NH3 + H20  NH4+ + OH-  water as an acid base acid H20 + H20  H30+ + OH-  water as an acid pure water has H2O, H3O+, OH- & base

(refer to your previous notes!)

(refer to your previous notes (refer to your previous notes! Make sure you have those particular ones memorized, both their formula and their name!)

(refer to your previous notes!)

Hydronium H3O+ Hydroxide OH- A hydrogen ion in water H+ + H2O  H3O+ H+ and H3O+ used interchangeably For acids Hydroxide OH- For bases

a measure of the number of hydronium or hydroxide ions is the pH scale it is based on the concentration of hydrogen and hydroxide ions in solution pH is defined as the negative logarithm of the hydrogen ion concentration

pH = -log([H+]) or -log([H3O+]) pOH = -log ([OH-]) [ ] = concentration (molarity) notice pH is based on a log scale log(base 10) is the power to which 10 must be raised to equal that number.

b n = x n = logb x Logarithm is an exponent 23 = 8 3 = log28 Then, 3 is called the logarithm of 8 with a base 2 3 = log28

104 = 10,000 4 = log1010,000 We’ll use this concept for pH The logarithm of 10,000 with a base 10 is 4 4 = log1010,000 We’ll use this concept for pH

14 = pH + pOH

0.001 M HCl In calculator: -log(0.001) (some calculators are a little different) pH = 3

Because NaOH is a base, we’re calculating the pOH 0.01 M NaOH Because NaOH is a base, we’re calculating the pOH pOH = -log(0.01) pOH = 2 14 = pH + pOH 2 pH = 12

3.2 x 10-9 M HNO3 pH = -log(3.2 x 10-9) pH = 8.49

the typical pH scale runs from 0 to 14 .

The concentration of an acid (or base) in solution can be determined by performing a neutralization reaction. Recall neutralization…

The reaction of an acid with a base produces water and a class of compounds called salts. HA + BOH  B A + H2O ex. HCl(aq) + NaOH(aq)  NaCl (aq) + H2O(l) acid base salt water

An indicator is used to show when neutralization has occurred. An indicator is a substance that forms different colors in different pH solutions. Phenolphthalein is a common indicator used in acid-base titrations. It will change from colorless in acidic environments to pink in basic environments.

A measured volume of a solution of unknown concentration (acid or base) is added to an Erlenmeyer flask. A solution of known molarity (acid or base) is added to a buret.

3. Several drops of an indicator are added to the unknown solution 4. Measured volumes of a solution of known molarity (acid or base) are mixed into unknown solution until the indicator just barely changes to a different color.

The solution of known concentration is called the standard solution The solution of known concentration is called the standard solution. The standard solution is added using a buret. The process of adding a known amount of solution of known concentration to determine the concentration of another solution is called titration. The point at which the indicator changes color is the end point of the titration.

This can also be done less “formally” using any volume measures This can also be done less “formally” using any volume measures. The results won’t be as accurate, but it gets you close. Example: (drops, substitute for mL)

aMAVA = bMBVB (similar to dilutions) Remember, in order for the solution to be neutral, Moles H30+ = moles OH- a = # of H+ ions in acids b = # of OH- ions in bases M = Molarity (M) V = volume aMAVA = bMBVB (similar to dilutions)

a = __ for H2SO4 and b = __ for Ca(OH)2 2 2 1) Determine a and b for the following a = __ for H2SO4 and b = __ for Ca(OH)2 2 2

MA = 0.749 M 1(MA)(35.28 mL) = 1(1.008M)(26.23 mL) 2) It takes 26.23 mL of a 1.008 M NaOH solution to neutralize 35.28 mL of a monoprotic acid solution. What is its molarity? 1(MA)(35.28 mL) = 1(1.008M)(26.23 mL) MA = 0.749 M

MB = 1.61 M 2(0.533 M)(23.40 mL) = 1(MB)(15.50 mL) 3) If 15.50 mL of NaOH solution were neutralized with 23.40 mL of 0.533 M H2SO4, what is the concentration of the NaOH ? 2(0.533 M)(23.40 mL) = 1(MB)(15.50 mL) MB = 1.61 M

4) 25.00 mL of 0.720 M nitric acid is used to completely neutralize a 1.0 M NaOH solution. What volume of NaOH is present? 1(0.720 M)(25.00 mL) = 1(1.0 M)VB VB = 18.00 mL

Ex: A 25 mL solution of nitric acid is completely neutralized by 18mL of 1.0M NaOH. What is the concentration of the acid solution?

Ex: A 25 mL solution of nitric acid is completely neutralized by 18 mL of 1.0M NaOH. What is the concentration of the acid solution? MA = ?? MB = 1M VA = 25mL VB = 18mL MAVA = MBVB rearranged to solve for MA MA= MBVB VA (1M)(18mL) = 0.720M 25mL

What is the molarity of phosphoric acid if 15 What is the molarity of phosphoric acid if 15.0 mL of the solution is completely neutralized by 38.5 mL of 0.150 M Al(OH)3? It takes 26.23 mL of a 1.32 x 10-3 M NaOH solution to neutralize a 45.56 mL of a triprotic acid solution. What is the concentration of the acid? 0.385 M H3PO4 MB = 2.53 x 104 M

3. What is the molarity of sodium hydroxide if 20 3. What is the molarity of sodium hydroxide if 20.0 mL of the solution is neutralized by 28.0 mL of 0.60 M HCl? 4. How many mL of 0.45 M HCl must be added to 25.00 mL of 1.00 M Sr(OH)2 to make a neutral solution? 0.840 M NaOH VA = 111.00 mL

What is the molarity of phosphoric acid if 15 What is the molarity of phosphoric acid if 15.0 mL of the solution is completely neutralized by 38.5 mL of 0.150 M Al(OH)3? H3PO4 Al(OH)3 MA = ?? MB = 0.150M VA = 15mL VB = 38.5mL MAVA = MBVB rearranged to solve for MA MA= MBVB VA (0.150M)(38.5mL) = 0.385M H3PO4 15mL

3. What is the molarity of sodium hydroxide if 20 3. What is the molarity of sodium hydroxide if 20.0 mL of the solution is neutralized by 28.0 mL of 0.60 M HCl? HCl NaOH MA = 0.60M MB = ?? VA = 28.0mL VB = 20.0mL MAVA = MBVB rearranged to solve for MB MB= MAVA VB (0.60M)(28.0mL) = 0.840M NaOH 20.0mL