Chapter 14 Acids and Bases. Naming Acids 2 types of acids Binary ternary (sometimes called oxy) ...

Chapter 14

Acids and Bases

Naming Acids

2 types of acids Binary ternary (sometimes called oxy)

binary -H and one other type of atom name them hydro _________ ic acid

Naming Acids

Ex1 HCl Hydrochloric Acid

Ex2 HBr Hydrobromic Acid

Ex3 H3P Hydrophosphoric Acid

Writing formulas from names for Acids

Criss Cross charges Ex4 Hydronitric Acid

H3N

Ex5 Hydrosulfuric Acid H2S

Naming Acids

ternary (oxy) acids H with a polyatomic ion Do not start with Hydro-

Change the –ate ending to –ic Change the – ite ending to –ous

Naming Acids

Ex6 H2SO4

Sulfuric Acid

Ex7 H2SO3

Sulfurous Acid

Ex8 HClO4 Perchloric Acid

Ex9 HClO Hypochlorous Acid

Writing Formulas From Names

Ex10 Nitric Acid HNO3

Ex11 Phosphorous Acid H3PO3

Some common acids:

Sulfuric – used for fertilizer, petroleum, production of metal, paper, paint

HCl – stomach acid, food processing, iron, steel

Acetic acid – vinegar, fungicide, produced by fermentation

Nitric acid – explosives, rubber, plastics, dyes, drugs

Phosphoric acid – beverage flavoring, animal feed, detergents

Properties of Acids:

Acid comes from Latin meaning acidus, or sour tasting.

Affect the colors of indicators. An indicator is a chemical that shows one color in an acid and another in a base. Acids turn blue litmus red.

Properties of Acids:

Acids react with bases to produce salt and water. This is called neutralization.

HCl(aq) + NaOH(aq) NaCl(aq) + HOH (l)

3H2SO4(aq)+ 2Al(OH)3(aq) Al2(SO4)3(aq) + 6 HOH (aq)

Properties of Acids:

Acids ionize in water. So, they conduct electricity (electrolytes).

Acids react with active metals to produce salts and hydrogen.

Mg+ + 2 HCl (aq) MgCl 2(aq) + H 2(g)

Cu(s) + HCl(aq) NR

Arrhenius Acids

Substances that produces H+ ions when mixed with water. HCl(g) + H2O(l) H+1

(aq) + Cl-1 (aq)

It is now found that: H+1

(aq) + H2O(l) H3O+1 (aq)

so it is really… HCl(g) + H2O(l) H3O+1

(aq) + Cl-1 (aq)

Definitions of Acids

Bronsted-Lowery Acids – proton donors

Show HCl + water and HCl + ammonia

HCl + H2O H3O+(aq) + Cl-1(aq)

HCl + NH3 NH4+

(aq) + Cl-1

(aq) (*not Cl2!!)

Types of Acids

Strong - HI, HBr, HCl, HNO3, H2SO4, HClO4 (one way arrows always!) HBr + H2O H3O+

(aq)+ Br-1 (aq)

Weak – HF, H2PO4, H2CO3, H2PO4

(double arrows always!) HF (aq) + H2O (l) H3O+1

(aq) + F-1 (aq)

Molecular, Total Ionic, and Net Ionic Equations for acids:

Molecular Equation: Zn(s) + 2 HCl(aq) ZnCl2(aq) + H 2(g)

Total Ionic Equation: Zn(s) + 2H+1

(aq) + 2Cl-1 (aq)

Zn+2(aq) + 2Cl-1(aq) + H2(g)

Net Ionic Equation: Zn(s) + 2H+1

(aq) Zn+2(aq) + H2(g)

Some acids donate more than 1 proton….

Monoprotic (HF) - an acid that donates one proton (one hydrogen) Ex1: Write the reaction(s) showing

the complete ionization of HF. HF (aq) + H2O (l) H3O+1

(aq) + F-1 (aq)

Some acids donate more than 1 proton….

Diprotic (H2SO4) - an acid that donates two protons (two hydrogens) Ex2: Write the reaction(s) showing the

complete ionization of H2SO4. H2SO4 (aq) + H2O (l) H3O+1 (aq) + HSO4

-1 (aq)

HSO4-1

(aq) + H2O(l) H3O+1(aq) + SO4

-2 (aq)

__________________________________H2SO4 (aq) + 2 H2O(l) 2 H3O +1

(aq)+SO4-2

(aq)

*Note: When you lose a H+1, you gain a negative.

Some acids donate more than 1 proton….

Triprotic Acid: an acid that donates three protons (three hydrogens).

Ex3: Write the reaction(s) showing the complete ionization of H3PO4.

H3PO 4(aq) + H2O (l) H3O +1 (aq) + H2PO4 -1 (aq)

H2PO4 -1 (aq) + H2O (l) H3O +1 (aq) + HPO4

-2 (aq)

HPO4 -2

(aq) + H2O (l) H3O +1 (aq) + PO4 -3 (aq)

________________________________________H3PO 4(aq) + 3 H2O (l) 3 H3O +1 (aq) + PO4

-3 (aq)

Some acids donate more than 1 proton….

Diprotic and Triprotic can also be referred to as polyprotic.

2nd and 3rd ionizations are always weak (so, ).

Bases

Bases are used in cleaners (floors, drains, ovens), react with fats and oils so they become water soluble, used to neutralize stomach acid (antacids), used as laxatives

Properties of Bases

Bases are electrolytes. They dissociate in water. NaOH and KOH are strong electrolytes because they are both highly soluble.

  Affect the colors of indicators. An indicator is a

chemical that shows one color in an acid and another in a base. Bases turn red litmus blue.

Bases react with acids to produce salt and water. This is called neutralization.

  Bases taste bitter and feel slippery. Soap is an

example of a base.

Definition of Bases

A substance that has OH- ions. Bases dissociate in water to give OH- & positive metal ions.

Types of Bases

1. Traditional Bases (Arrhenius) – a substance that contains hydroxide ions and dissociates to give hydroxide ions in water.

NaOH(s) + H2O Na+(aq) + OH-

(aq)

Mg(OH)2(s) + H2O Mg+2(aq) + 2OH-

(aq)

Types of Bases

2. Bronsted- Lowry bases – proton acceptors 

NH3(g) + H2O(l) NH4+1

(aq) + OH-1(aq)

* Water is amphoteric. It can act as an acid

or base.

Types of Bases

List of strong bases: NaOH, KOH, CsOH, Ca(OH)2

List of weak bases: many organic compounds with N…NH3,

C6H5NH2, C2H3O2-

Hydroxides of Column I and II are strong bases

Neutralization reactions – hydronium + hydroxide yields water

It is a type of double replacement reaction. Note: H2O = HOH Acid + Base → Salt and Water

General Formula: HX + MOH MX + H2O

Neutralization Reaction

Example: hydrochloric acid + barium hydroxide ( molecular, total ionic, net ionic)

2HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2HOH(l)

2H+1(aq) + 2Cl-1(aq) + Ba+2

(aq) + 2OH-1(aq)

Ba+2(aq) + 2Cl-1(aq) + 2HOH(l)

2H+1(aq) + 2OH-1

(aq) 2HOH(l)

Relative Strengths of Acids and Bases

Some acids and bases are stronger than others. Bronsted (Danish) and Lowry (English)

independently discovered that acids are proton donors and bases are proton acceptors. A proton is a hydrogen ion.

Relative Strengths of Acids and Bases

Strong Acid Example: HCl(g) + H2O(l) H3O+1

(aq) + Cl-1(aq)

(Acid) (Base)

Weak Base Example: NH3(g)+ H2O(l) NH4

+1(aq) + OH-1

(aq)

(Base) (Acid)

Remember – water is amphoteric!

Conjugate Acids & Bases

Conjugate Acid: the substance that was the base and now acts as an acid.

Conjugate Base: the substance that was the acid and now acts as a base.

HCl(g) + H2O(l) H3O+1(aq) + Cl-1(aq)

(Acid) (Base) (Conjugate Acid) (Conjugate Base)

Conjugate Acids & Bases

NH3(g) + H2O(l) NH4+1

(aq) + OH-1(aq)

  HF(l) + H2O(l) H3O+1

(aq) + F-1(aq)

Conjugate Acids and bases

H2CO3(aq) + H2O(l) H3O+1(aq) + HCO3

-1(aq)

The stronger the acid, the weaker the conjugate base.

Proton transfer reactions favor the production of the weaker acid and the

weaker base.

Conj. Acid/Base Practice

Complete the equation and label acid base pairs HSO4

-1(aq) + HCO3

-1(aq)

Write an equation showing how NH2-1 is a

stronger base than HSO4-1

Conj. Acid/Base Practice

Which one is correct? HSO4

-1(aq) + H3O+1

(aq) H2SO4(aq) + H2O(l)

  (Base) (Acid) (Conj. Acid) (Conj. Base)

or HSO4

-1(aq) + OH-1

(aq) SO4-2

(aq) + H2O(l)

(Acid) (Base) (Conj. Base) (Conj. Acid)

The second reaction is favored because a weaker conjugate acid/base is produced.

Stuff to know for Acids and Bases

2nd and 3rd ionizations are always weak. This means a double yield sign ().

Memorize these strong acids. Strong means a single yield sign ().

HI, HBr, HCl, HNO3, H2SO4, HClO4

All other acids get double yield signs. Strong bases include metals from

column #1 and column #2 (below magnesium).

Proton reactions favor the formation of the weaker acid and base.

Reactions of Acids and Bases

Neutralization (double replacement): Acid + Base Salt + Water HX + MOH MX + H2O

Ex1: HCl(aq) + NaOH(aq) NaCl(aq) + HOH(l)

Reactions of Acids and Baes

Acid + Metal (single replacement): Metal + Acid Salt + Hydrogen

Ex2: Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)

Reactions of Acids and Bases

Acid in water: Acid + Water Hydronium Ion + Negative Ion

Ex3: HCl(g) + H2O(l) H3O+1(aq) + Cl-1(aq)

Reactions of Acids and Bases

Traditional Base (ends with OH) in water (dissociation): Base + Water Positive Ion + Hydroxide

Ex4: Fe(OH)3(s) + H2O(l) Fe+3

(aq) + 3 OH-1(aq)

Reactions of Acids and Bases

Formation of acids and bases from anhydrides - synthesis (anhydride “without water”): Nonmetal oxide + water acid Ex5a:

CO2(g) + H2O(l) H2CO3(aq)

Ex5b: SO3(g) + H2O(l) H2SO4(aq)

Note: just add the nonmetal oxide to the water to determine the product.

Reactions of Acids and Bases

Metal oxide + water base Ex5c: Na2O(s) + H2O(l) 2 NaOH(aq)

Ex5d: MgO(s) + H2O(l) Mg(OH)2(aq)

Reactions of Acids and Bases

Acid and metal oxide (really just an acid and a base): Acid + Metal Oxide Salt + Water Ex6: H2SO4(aq) + CuO(s)

Turn CuO into Cu(OH)2

H2SO4(aq) + Cu(OH)2 (aq) CuSO4(aq) + HOH(l)

Now, re-write the original reactants, new products, and balance.

H2SO4(aq) + CuO(s) CuSO4(aq) + H2O(l)

Reactions of Acids and Bases

Base and nonmetal oxide (really just an acid and a base Base + Nonmetal Oxide Salt + Water

Ex7a: CO2(g) + NaOH(aq) NaHCO3(aq)

Reactions of Acids and Bases

This is a little confusing. So these reactions will be done like: CO2(g) + NaOH(aq)

Turn CO2 into H2CO3

H2CO3(aq) + NaOH(aq) Na2CO3(aq) + HOH(l)

Now, re-write the original reactants, new products, and balance. CO2(g) + 2 NaOH(aq) Na2CO3(aq) + H2O(l)

Reactions of Acids and Bases

Ex7b: 2 CO2(g) + Ca(OH)2(aq) Ca(HCO3)2(aq)

CO2(g) + Ca(OH)2(aq) Turn CO2 into H2CO3

H2CO3(aq) + Ca(OH)2(aq) CaCO3(aq) + HOH(l)

Now, re-write the original reactants, new

products, and balance. CO2(g) + Ca(OH)2(aq) CaCO3(aq) + H2O(l)

Reactions of Acids and Bases

Metal oxide and nonmetal oxide It is like an acid base reaction they yield salt.

However, it does not produce water since no hydrogen is involved.

Ex8a: MgO(s) + CO2(g) MgCO3(s)

Note: just add the nonmetal oxide to the metal oxide to determine the product. Ex8b: CuO(s) + SO3(g) CuSO4(s)

Aqueous Solutions and the Concept of pH

Tap water conducts electricity – why? – many ions present: examples: Distilled water appears to not conduct

electricity, but it does – just a little, tiny bit

H2O + H2O H3O+1 + OH-1

Aqueous Solutions and the Concept of pH

The normal way to express the quantity of hydronium and hydroxide ions is in moles/L (M)

At 25 C0, [H3O+1] = 1 x 10-7 so [OH-1] = 1 x 10-7

These numbers are constant in neutral solution, so we can multiply them to get a constant

We call this constant Kw - ionization constant for water

Kw = [H3O+1][OH-1]

Aqueous Solutions and the Concept of pH

At 25 C0, [H3O+1] = 1 x 10-7

so [OH-1] = 1 x 10-7

so Kw = 1 x 10-14

Example: If the [H3O+1] is 1 x 10-3M, then what is the [OH-1]?

The solution is acidic because the hydronium ion concentration is greater than the hydroxide concentration.

Kw Practice

Beaker # [H3O+1] [OH-1] Acid or Base

1 1 x 10-5 1 x 10-9 Acid

2 1 x 10-12 1 x 10-2 Base

3 2 x 10-4 5 x 10-11 Acid

4 2.40 x 10-9 4.16 x 10-6 Base

Fill in the table below:

Aqueous Solutions and the Concept of pH

pH stands for parts per million of

Hydrogen ion.

How to calculate strengths of Acids and Bases

pH = - log 10 [H3O+]

pOH = - log 10 [OH-]

[H3O+] = 10 –pH

[OH-] = 10-pOH

pH + pOH = 14

How to calculate strengths of Acids and Bases

Log Review Logs are functions of exponents ex. log of 1000 = ex. log of .01 =

How to calculate strengths of Acids and Bases

To convert [H3O+1] to pH pH = - log 10 [H3O+] log, #, enter, then make it positive –

change the sign) [H3O+1] = pH =

1.0 x10 –1  

1.0 x 10 –2  

3.0 x 10 –4  

How to calculate strengths of Acids and Bases

To convert pH to [H3O+1] [H3O+] = 10 –pH

2nd, log, (-), #, enter

[H3O+1] pH Acid or Base

2

11

5.22

How to calculate strengths of Acids and Bases

[H+] [OH-] pH pOH Acid or Base

2 x 10-5

3.5 x 10-5

3.25

8.12

8.20

0.0016

2.8 x 10-11

Concentration units for Acids and Bases

Chemical Equivalents: quantities of solutes that have equivalent combining capactieies.

Ex1: HCl + NaOH NaCl + H2O

To achieve a balance, 1 mole H+ needs to cancel out 1 mole OH-1.

So, for the above reaction: 1 mole of HCl is necessary to balance 1 mole of NaOH.

1 mole HCl = 1 mole NaOH

Concentration units for Acids and Bases

Ex2: H2SO4 + 2 NaOH Na2SO4 + 2 H2O

To achieve a balance, 1 mole H+ needs to cancel out 1 mole OH-1.

So, for the above reaction: 1/2 mole of H2SO4 is necessary to balance 1 mole of NaOH.

½ mole H2SO4 = 1 mole NaOH

Concentration units for Acids and Bases

Ex3: To make H3PO4 chemically equivalent to NaOH, 1/3 mole of H3PO4 balances 1 mole NaOH

Equivalent Weight

the # of grams of acid or base that will provide 1 mole of protons or hydroxide ions.

HCl H2SO4 H3PO4

Moles of Acid 1 ½ 1/3

Moles of Hydrogen

1 1 1

Equivalent Weights

36.5 49.05 32.7

Equivalent Weight

Formula for calculating equivalent weight Eq. wt. = MW / equivalents

MW = molecular weight

Formula for calculating equivalents: # equivalents = (moles)(n)

n = # of H or OH in the chemical formula

Calculations

Example 1: How many equivalents in 9.30 g of H2CO3?

Step 1: Calculate the molecular weight of H2CO3:

H = (2)1.0 = 2.0C = (1)12.0 = 12.0O = (3)16.0 = 48.0

mw = 62.0 g/mole 

Calculations

Step 2: Calculate the # of moles in 9.30 g of H2CO3: 9.30 g H2CO3 x 1 mole H2CO3 = .150 mole

62.0 g H2CO3

H2CO3

  Step 3: Calculate the number of equivalents

# equivalents = (moles)(n) n = # of H or OH in the chemical formula

eq = (.150 moles)(2) eq = .300 eq

Calculations

Ex 2: Calculate the equivalent weight of 9.30 g H2CO3 (Use 00.300 equivalents calculated

above)eq. wt. = mw/eq

eq. wt. = 62.0 g / .300 equivalents eq. wt. = 206.7 g/eq

Calculations

Ex 3: How many grams of H2CO3 would equal .290 equivalents?

Step 1: Convert to moles # equivalents = (moles)(n)

moles = .29 / 2 moles = .145 moles

Step 2: convert moles to grams: .145 moles H2CO3 x 62.0 g H2CO3 = 8.99 g

1 mole H2CO3 H2CO3

Normality

In the past, we have used M for concentration. M = moles / L A more useful form of concentration for

acid/base reactions is Normality. N = # eq / L

# equivalents = (moles)(n) n = # of H or OH in the chemical formula

Normality

Also, in calculating pH, normality is used over molarity.

Normality is related to Molarity: N = (M)(n)

M = moles/liters (total) n = # of H or OH in the chemical formula

Normality

Ex 1: Find the normality of a solution that contains 1 mole H2SO4 in 1 L solution

Step 1: Calculate the molarity:1 mole H2SO4 = 1 M

1 L Step 2: Calculate the normality: N = (M)(n)

N = 1 M x 2N = 2 N

Normality

Ex2: Calculate the Normality if 1.80 g of H2C2O4 is dissolved in 150 mL of solution.

Step 1: convert grams to moles: 1.80 g H2C2O4 x 1 mole H2C2O4 = .0200 moles

90.0 g H2C2O4 H2C2O4

Step 2: Calculate the molarity: M = .0200 moles / .150 L .133 M Step 3: Calculate the normality:

N = (M)(n) N = (.133)(2) .267 N

Problems involving mixing unequal amounts of acid and base

Ex1: Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.

Step 1: Find the moles of HCl and moles of NaOH:HCl .100 M = x / .0500 L x

= .00500 moles HCl

NaOH .100 M = x / .0490 L x =.00490 moles NaOH

Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.

Step 2: Find the equivalents of H+ in HCl and the equivalents of OH- in NaOH:

eq = (moles)(n)

H+ (.00500 moles)(1) = .00500 eq of H+

OH- (.00490 moles)(1) = .00490 eq of OH-

Step 3: Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other (absolute value):

.00500 eq H+ - .00490 eq OH- = .00010 eq H+

Find the pH of a solution made by mixing 50.0 mL of .100 M HCl with 49.0 mL of .100 M NaOH.

Step 4:Calculate the normality of the resulting solution: N = eq / L

Total liters 50.0 mL + 49.0 mL = 99.0 mL .099 L

N = .00010 eq / .0990 L .00101 N

Step 5: Calculate the pH: pH = - log [H+] pH = - log (.00101) pH = 3.00

Example 2:

Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH

Step 1: Find the moles of H2SO4 and moles of NaOH

Step 2: Find the equivalents of H+ in H2SO4 and the equivalents of OH- in NaOH

Step 3: Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other

Step 4: Calculate the normality of the resulting solution

Step 5: Calculate the pH

Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH

Find the moles of H2SO4 and moles of NaOH:

H2SO4 .100 M = x / .0500 L x = .00500

moles H2SO4

NaOH 1.00 M = x / .0500 L x = .0500 moles NaOH

Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH

Find the equivalents of H+ in H2SO4 and the equivalents of OH- in NaOH:

eq = (moles)(n)H+ (.00500 moles)(2) = .0100 eq

OH- (.0500 moles)(1) = .0500 eq

Find the equivalents of H+ or OH- left over by subtracting the equivalents of H+ and equivalents of OH- from each other (absolute value):

.0500 eq OH-1 - .0100 eq H+ = .0400 eq OH-

Find the pH when mixing 50.0 mL of .100 M sulfuric acid with 50.0 mL of 1.00 M NaOH

Calculate the normality of the resulting solution:N = eq / L

total L 50.0 mL + 50.0 mL = 100.0 mL .1000L

N = .0400 eq / .1000 L .400 N

Calculate the pH: 1st calculate the pOH: pOH = -log[OH-] pOH = -log(.400) pOH =.399 Calculate the pH from the pOH pH = 14 – pOH pH = 14 - .399 pH = 13.6

Titrations

A controlled addition and measurement of the amount of a solution of a known concentration that is required to react completely with a measured amount of a solution of unknown concentration.

Titrations

Titrations

Equivalence point – In a neutralization reaction, the point @ which there are equivalent quantities of H3O+ and OH-

End Point – point in a titration where the indicator changes color

Titration of 20 mL 0.1 M HCL

mL of 0.1 M NaOH Titrant

Titration of HCl with NaOH

High amount of H+1

H+1= OH-

High amount of H+1

End point

Normality

Normality and Titration: Equation: NaVa =NbVb

N = normality V = volume in Liters

Ex1: A 15.5 mL sample of .215 M KOH requires 21.2 mL of acetic acid to titrate to the end point. What is the Molarity of the acid?

Step 1: Convert molarity to normality: N = (M)(n)

n = # of H or OH in the chemical formula N = (.215)(1) .215 N KOH

 Step 2: Use the titration formula to find the normality of the acid:

NaVa =NbVb

(x)(21.2 mL) = (.215 N)(15.5 mL) x = .157 N

Step 3: Convert normality to molarity:M = N / n x = .157 N / 1 x = .157 M

Ex2: If 15.7 mL of sulfuric acid is titrated to the end point by 17.4 mL of .0150 M NaOH, what is the Molarity of the acid?

Convert molarity to normality: N = (M)(n)

N = (.015)(1) .0150 N KOH

Use the titration formula to find the normality of the acid:

NaVa =NbVb

(x)(15.7 mL) = (.0150 N)(17.4 mL) x = .0166 N

Convert normality to molarity: M = N / n x = .0166 N / 2 x = .00831 M

Titration Problems

Ex 3: In a titration of 27.4 mL of 0.0154 M Ba(OH)2 solution is added to a 20 mL sample of an HCl solution. What is the Molarity of the HCl solution?

Percent Problems

Long way……. If 18.75 mL of .750 N NaOH is required to titrate

20.30 mL of acetic acid, calculate the % acetic acid in solution.

Step 1: Use the titration formula to find the normality of the acid:

NaVa =NbVb

(x)(20.30 mL) = (.750 N)(18.75 mL) x = .693 N

If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.

Step 2: Multiply the normality by the equivalent weight of the acid:

.693 N .693 eq/L

(.693 eq/L)(60.0g/eq) = 41.6 g/L

If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.

Step 3: Convert the liters to grams:

41.6 g/L 41.6 g/1000mL 41.6 g/1000g  Step 4: Calculate the % by mass: (41.6 g / 1000g) 100 = 4.16 %

If 18.75 mL of .750 N NaOH is required to titrate 20.30 mL of acetic acid, calculate the % acetic acid in solution.

The short way…… Use the titration formula to find the normality of the acid:

NaVa =NbVb

(x)(20.30 mL) = (.750 N)(18.75 mL) = x = .693 N

Use: (N)(eq wt)/10 = %

(.693)(60.0) / 10 = x

x = 4.16 %