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 Ex2 HBr Ex3 H3P Hydrochloric Acid 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 Ex7 H2SO3 Ex8 HClO4 Ex9 HClO Sulfuric Acid 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 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 * Water is amphoteric. It can act as an acid 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: List of weak bases: Hydroxides of Column I and II are strong bases List of strong bases: NaOH, KOH, CsOH, Ca(OH)2 List of weak bases: many organic compounds with N…NH3, C6H5NH2, C2H3O2-

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 Fill in the table below: 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 4 2.40 x 10-9 4.16 x 10-6

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 [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 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 C = (1)12.0 = 12.0 O = (3)16.0 = 48.0 mw = 62.0 g/mole Example 1: How many equivalents in 9.30 g of H2CO3? Step 1: Calculate the molecular weight of H2CO3: H = (2)1.0 = 2.0 C = (1)12.0 = 12.0 O = (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 2 N = 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 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 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 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 HCl with NaOH Titration of 20 mL 0.1 M HCL High amount of H+1 End point H+1= OH- High amount of H+1 mL of 0.1 M NaOH Titrant

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 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 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

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

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