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by Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.

Properties of Acids Sour taste Change color of vegetable dyes (“indicators”) React with “active” metals Like Al, Zn, Fe, but not Cu, Ag or Au Zn + 2 HCl  ZnCl 2 + H 2 Corrosive React with carbonates, producing CO 2 Marble, baking soda, chalk CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O React with bases to form ionic salts, and often water

Properties of Bases Also known as alkalis Bitter Taste Feel slippery Change color of vegetable dyes Different color than acid Litmus = blue React with acids to form ionic salts, and often water (HCl + NaOH  NaCl + HOH) Neutralization

Bubbles (CO 2 ) NR Limestone CaCO 3 Bubbles (H 2 ) NR Magnesium Yellow Blue Bromothymol Stayed clear (cloudy) Clear  Pink (Fuschia) Phenolphthalein Blue  Red Red  Blue Litmus (blue or red) <7 >7 pH (# from the key) Not slippery Slippery Feel (choose slippery or not slippery) Sour Bitter Taste Acids ( H + + A - ) Bases (X + + OH - ) Lab Results

Mnemonic Device H + A - B + OH - clea R O Y G B I V uschia 1 7 14 BTB universal phenolphthalein pH = litmus

Arrhenius Theory Acids ionize in water to H + ions and anions Bases ionize in water to OH - ions and cations Neutralization reaction involves H + combining with OH - to make water H + ions are protons

Definition only good in water solution Definition does not explain why ammonia solutions turn litmus blue Basic without OH - ions Arrhenius Theory (cont.)

Brønsted-Lowery Theory H + transfer reaction Since H + is a proton, also known as proton transfer reactions In the reaction, a proton from the acid molecule is transferred to the base molecule Products are called the conjugate acid and conjugate base

Brønsted-Lowery Theory (cont.) H-A + :B  A - + H-B + A - is the conjugate base, H-B + is the conjugate acid Conjugate acid-base pair is either the original acid and its conjugate base or the original base and its conjugate acid H-A and A - are a conjugate acid-base pair :B and H-B + are a conjugate acid-base pair

Complete the following table Produces H+ ions in solution Produces OH- ions in solution Proton donor Proton acceptor

Example #1: Determine what species you will get if you remove 1 H +1 from the acid. Conjugate base will have one more negative charge than the original acid H 3 PO 4  H + + H 2 PO 4 - Write the conjugate base for the acid H 3 PO 4

Brønsted-Lowery Theory (cont.) In this theory, instead of the acid, HA, dissociating into H + (aq) and A - (aq), the acid donates its H to a water molecule HA + H 2 O  A - + H 3 O + A -1 is the conjugate base H 3 O + is the conjugate acid

Brønsted-Lowery Theory (cont.) H 3 O + is called the hydronium ion In this theory, substances that do not have OH - ions can act as a base if they can accept a H +1 from water. H 2 O + :B  OH - + H-B +

Identify the acid and base in the following reaction acid base Conjugate acid (hydronium ion) Conjugate base Conjugate acid-base pair Conjugate acid-base pair

Bases in Water acid base Conjugate acid (hydronium ion) Conjugate base Conjugate acid-base pair Conjugate acid-base pair

Identify the conjugate acid-base pairs HSO 4 - + H 2 O  SO 4 2- + H 3 O + HCO 3 - + H 2 O  OH - + H 2 CO 3 HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2 - + H 3 O + HPO 4 2- + H 2 O  H 2 PO 4 - + OH -

Identify the conjugate acid-base pairs HSO 4 - + H 2 O  SO 4 2 - + H 3 O + A B CB CA HCO 3 - + H 2 O  OH - + H 2 CO 3 B CA CB CA HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2 - + H 3 O + A B CB CA HPO 4 2- + H 2 O  H 2 PO 4 - + OH - B A CA CB

Complete the following tables NO 3 - H 2 O OH - H 2 SO 4 Br - OH - H 3 O + HCO 3 - HSO 4 - ClO 4 -

Strength of Acids & Bases The stronger the acid, the more willing it is to donate H + (i.e. 100% dissociation)

Strength of Acids & Bases (cont.) Strong bases will react completely with water to form hydroxide: CO 3 -2 + H 2 O  HCO 3 - + OH - Only small fraction of weak base molecules pull H + off water: HCO 3 - + H 2 O  H 2 CO 3 + OH -

Multiprotic Acids Monoprotic acids have 1 acid H, diprotic 2, etc. In oxyacids only the H on the O is acidic (why?) In strong multiprotic acids, like H 2 SO 4 , only the first H is strong; transferring the second H is usually weak H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - HSO 4 - + H 2 O  H 3 O + + SO 4 -2

What new words describe H 2 SO 4 ?

What new words describe H 2 SO 4 ? Strong acid  100% dissociated (i.e. strong electrolyte) Diprotic (two protons that it can donate) Oxyacid (contains oxygen) Note: since it is a strong acid HSO 4 - is a weak base (strong acids always form weak conjugate bases)

Water As an Acid and a Base Amphoteric substances can act as either an acid or a base. Water as an acid, NH 3 + H 2 O  NH 4 + + OH - Water as a base, HCl + H 2 O  H 3 O + + Cl - Water can even react with itself: H 2 O + H 2 O  H 3 O + + OH -

Autoionization of Water Water is an extremely weak electrolyte. Therefore there must be a few ions present H 2 O + H 2 O  H 3 O + + OH -

* Acidic and Basic Solutions * Acidic solutions have a larger [H + ] than [OH - ] Basic solutions have a larger [OH - ] than [H + ] Neutral solutions have [H + ]=[OH - ]= 1 x 10 -7 M K w = [H + ][OH - ]= 1 x 10 -14 [H + ] = 1 x 10 -14 [OH - ] [OH - ] = 1 x 10 -14 [H + ]

Example #2 Determine the [H + ] and [OH - ] in a 10.0 M H + solution

Example #2 (cont.) Determine the given information and the information you need to find Given [H + ] = 10.0 M, find [OH - ]

Given [H + ] = 10.0 M = 1.00 x 10 1 M K w = 1.0 x 10 -14 Example #2 (cont.)

pH & pOH The acidity/basicity of a solution is often expressed as pH or pOH. pH = -log[H 3 O + ] pOH = -log[OH - ] pH water = -log[10 -7 ] = 7 = pOH water [H + ] = 10 -pH [OH - ] = 10 -pOH

pH & pOH (cont.) pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution 1 pH unit corresponds to a factor of 10 difference in acidity 14 = pH + pOH

Example #3 Calculate the pH of a solution with a [OH - ] = 1.0 x 10 -6 M

Example #3 (cont.) Find the concentration of [H + ]

Enter the [H + ] concentration into your calculator and press the log key log(1.0 x 10 -8 ) = -8.0 Change the sign to get the pH pH = -(-8.0) = 8.0 Example #3 (cont.)

Enter the [H + ] or [OH - ] concentration into your calculator and press the log key log(1.0 x 10 -3 ) = -3.0 Change the sign to get the pOH pOH = -(-3) = 3.0 Subtract the calculated pH or pOH from 14.00 to get the other value pH = 14.00 – 3.0 = 11.0 Example #4 Calculate the pH and pOH of a solution with a [OH - ] = 1.0 x 10 -3 M

If you want to calculate [OH - ] use pOH; if you want [H + ] use pH. It may be necessary to convert one to the other using 14 = [H + ] + [OH - ] pOH = 14.00 – 7.41 = 6.59 Example #5 Calculate the [OH - ] of a solution with a pH of 7.41

Example #5 (cont.) Enter the pH or pOH concentration into your calculator Change the sign of the pH or pOH -pOH = -(6.59) Press the button(s) on you calculator to take the inverse log or 10 x [OH - ] = 10 -6.59 = 2.6 x 10 -7 M

Calculating the pH of a Strong, Monoprotic Acid A strong acid will dissociate 100% HA  H + + A - Therefore the molarity of H + ions will be the same as the molarity of the acid Once the H + molarity is determined, the pH can be determined pH = -log[H + ]

Example #6 Calculate the pH of a 0.10 M HNO 3 solution. pH means – log of [H+] pH = - log [0.10] pH = - log [ 1 x 10 -1 ] pH = 1 note the exponent!

Example #6 (cont.) Determine the [H + ] from the acid concentration HNO 3  H + + NO 3 - 0.10 M HNO 3 = 0.10 M H + Enter the [H + ] concentration into your calculator and press the log key log(0.10) = -1.00 Change the sign to get the pH pH = -(-1.00) = 1.00

Practice Problems A HCl solution is 8.34 x 10 -5 mole/liter. Estimate, then calculate the pH of the solution. What is the [OH - ] of a solution whose pOH = 2.86 ? What is the [OH - ] of a solution whose [H + ] = 0.001M The pH of a soft drink is determined to be 4.0. What is the [OH - ] of the drink? What is the pH of a 0.001 M Mg(OH) 2 solution? (Assume 100% dissociation)

Practice Problems A HCl solution is 8.34 x 10 -5 mole/liter. Estimate, then calculate the pH of the solution. pH < 5 (see the exponent) pH = - log [H+] pH = - log[8.34 x 10 -5 ] pH = 4.079 (pH < 7, acidic)

Practice Problems What is the [OH - ] of a solution whose pOH = 2.86 ? pOH means – log[OH-] -log[OH-] = 2.86 log[OH-] = - 2.86 [OH-] = 10 -2.86 [OH-] = 0.0014 M since [OH-]>[H+] solution is basic

Practice Problems What is the [OH - ] of a solution whose [H + ] = 0.001M [H+][OH-] = 1 x 10 -14 [0.001][OH-] = 1 x 10 -14 [OH-] = 1 x 10 -11 Since [OH-] < [H+] solution is acidic

Practice Problems The pH of a soft drink is determined to be 4.0. What is the [OH - ] of the drink? pH + pOH = 14 4 + pOH = 14 pOH = 10 -log [OH] = 10 [OH-] = 10 -10 since [OH-] < [H+] soda is acidic

Practice Problems What is the pH of a 0.001 M Mg(OH) 2 solution? (Assume 100% dissociation) [OH-] = 2 * 0.001 pOH = - log [0.002] pOH = 2.699 pH + pOH = 14 pH = 14 – 2.699 = 11.30 (pH > 7, basic)

Challenge Questions What is the pOH of a 0.0025 M acetic acid solution that is only 8.5% dissociated? How much more acidic is a solution whose pH is 6.0 compared to a solution whose pH is 12.0? What is the resulting pH if equal volumes of solutions are mixed, one with a pH of 6.0 and one with a pH of 12.0?

Challenge Questions What is the pOH of a 0.0025 M acetic acid solution that is only 8.5% dissociated? HA  H+ + A- 0.0025 0 0 - 2.125 x 10 -4 + 2.125 x 10 -4 +2.125 x 10 -4 0.0025 * 0.085 = 2.125 x 10 -4 M [H+] 14 – (-log(2.125 x 10 -4 ) =

Challenge Questions How much more acidic is a solution whose pH is 6.0 compared to a solution whose pH is 12.0? pH = 6, [H+] = 0.000001 pH = 12 [H+] = 0.000000000001

Challenge Questions What is the resulting pH if equal volumes of solutions are mixed, one with a pH of 6.0 and one with a pH of 12.0? pH = 6, [H+] = 0.000001 pH = 12 [H+] = 0.000000000001 (0.000001 + 0.000000000001)/2 = - log ( ) =

Titration Laboratory Set-up Sample Problem Determine the unknown concentration of HCl if 25.0 mL of the acid are neutralized with 50.0 mL of 0.100 M NaOH. H + + OH -  HOH

Practice Problems Calculate the volume of 0.300 M HCl needed to titrate 75.00 mL of 0.1500 M KOH (aq) . Determine the volume of 0.100 M NaOH needed to reach the equivalence (end) point against 50.0 mL of 0.200 M HNO 3.

Buffered Solutions Buffered solutions resist change in pH when an acid or base is added to it. Used when need to maintain a certain pH in the system Blood

Buffered Solutions (cont.) A buffer solution contains a weak acid and its conjugate base. Buffers work by reacting with added H + or OH - ions so they do not accumulate and change the pH. Buffers will only work as long as there are sufficient weak acid and conjugate base molecules present.

Challenge Question Give two components of a buffer. Identify which component will react with added acid, show using a balanced equation. Identify which component will react with added base, show using a balanced equation.

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Zum notes (09) acids and bases

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Zum notes (09) acids and bases

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