Gen Chem II > Chapter 15-16 > Flashcards

Chapter 15-16 Flashcards

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1
Q

A weaker bond on an acid makes it more or less acidic

A

More acidic

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2
Q

The pronunciation for, HNO3

A

Nitric acid

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3
Q

THERE ARE TWO WAYS TO MAKE A BUFFER

A

SIGNIFICANT AMOUNTS OF A WEAK ACID AND IT’S CJB OR WEAK BASE AND ITS CJA

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4
Q

BUFFERS: IF A STRONG BASE (NAC2H3O2) IS ADDED, IT WILL BE NEUTRALIZED BY THE:

A. STRONG ACID
B. WEAK ACID

A

B. WEAK ACID

* (HC2H3O2) 
                      NAOH + HC2H3O2 --> H20 + NAC2H3O2
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5
Q

NAOH + HC2H3O2 –> H20 + NAC2H3O2

IF THE AMOUNT NaOH ADDED IS LESS THAN THE AMOUNT OF ACETIC ACID PRESENT IN THE BUFFER, THE pH CHANGE IS:

A. LARGE
B. SMALL

A

B. SMALL

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6
Q

THE BUFFERING ACTION OF THE ACID CAN BE UNDERSTOOD BY APPLYING WHO’S PRINCIPLE TO WEAK ACID EQUILIBRIUM

A

LE CHATELIER’S

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7
Q

HOW TO BUFFER ACID ADDITIONS:

IF A STRONG ACID IS ADDED, IT IS NEUTRALIZED BY THE:

A. CJA IN THE BUFFER
B. CJB IN THE BUFFER

A

B. CJB IN THE BUFFER (C2H3O2-)

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8
Q

WOULD, A SOLUTION THAT IS 0.100 M HNO2 AND 0.100 M HCl, MAKE A BUFFER

A

NO.

HNO2: WEAK ACID
HCl: STRONG ACID

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9
Q

WOULD, A SOLUTION THAT IS 0.100 M HF AND 0.100 M NaF, MAKE A BUFFER

A

YES. (HAS BASIC CJ PAIR)

HF: WEAK ACID
NaF: CJB

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10
Q

WOULD, A SOLUTION THAT IS 0.100 M HNO2 AND 0.100 M NaCl, MAKE A BUFFER

A

NO.

HNO2: WEAK ACID
NaCl: WEAK BASE

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11
Q

WHICH BASE IS WEAKEST:

NaCl OR NaOH

A

NaCl

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12
Q

STRONG ACIDS COMBINE WITH STRONG BASES TO FORM:

A

WEAKER ACIDS AND BASES

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13
Q

WOULD, A SOLUTION THAT IS 0.100 M HNO2 AND 0.100 M NaNO2, MAKE A BUFFER

A

YES.

HNO2: WEAK ACID
NaNO2: CJB

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14
Q

WOULD, A SOLUTION THAT IS 0.100 M HCl AND 0.100 M NaCl, MAKE A BUFFER

A

NO.

HCl: STRONG ACID
NaCl: WEAK BASE

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15
Q

MADE BY ADDING A WEAK ACID AND A SALT CONTAINING THE CJB OF A WEAK ACID

A

A BUFFER.

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16
Q

NaA, IS THE CJB OF A:

A: STRONG ACID
B: WEAK ACID

A

B. WEAK ACID

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17
Q

AN EQUATION DERIVED FROM THE Ka EXPRESSION THAT ALLOWS US TO CALCULATE THE pH OF A BUFFER SOLN. IS THE:

A: LE CHATELIER’S PRINCIPLE
B. HENDERSON-HASSELBALCH EQUATION

A

B. HENDERSON-HASSELBALCH EQUATION

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18
Q

SOLN. IS SATURATED, NO PRECIPITATION:

A. Q Ksp
C. Q= Ksp

A

C. Q= Ksp

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19
Q

SOLN. IS UNSATURATED, NO PRECIPITATION:

A. Q Ksp
C. Q= Ksp

A

A. Q

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20
Q

SOLN. IS SUPERSATURATED, PRECIPITATION OCCURS:

A. Q Ksp
C. Q= Ksp

A

B. Q > Ksp

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21
Q

THE LOWER THE pH:

A. THE HIGHER THE SOLUBILITY
B. THE LOWER THE SOLUBILITY

A

A. THE HIGHER THE SOLUBILITY

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22
Q

SOLUBILITY PRODUCT (FOR THE DISSOCIATION OF (S) SALT INTO (aq) IONS

A

Ksp

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23
Q

THE TWO WAYS BUFFERS CAN BE MADE BY

A
  1. MIXING WEAK ACID AND IT’S CJB

2. MIXING WEAK BASE AND IT’S CJA

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24
Q

IN HENDERSON-HASSELBALCH EQUATION, THE CHEMICAL RXN IS WRITTEN C A WEAK BASE AS A REACTANT AND ITS CJA AS A PRODUCT:

B: + H2O __?_ + __?_

A

B: + H2O H:B+ + OH-

Kb = [H:B+][OH-]
—————-
[B:]

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25
IN HENDERSON-HASSELBALCH EQUATION, THE CHEMICAL RXN IS WRITTEN C A WEAK ACID AS A REACTANT AND ITS CJB AS A PRODUCT: H:B+ + H2O B: + H3O+ IDENTIFY THE WEAK ACID AND CJB:
(H:B+) IS THE WEAK ACID, (B:) IS THE CJB [B:] pH = pKa + log ( ------ ) [H:B+]
26
BUFFERS RESIST CHANGE IN pH, BUT WHAT HAPPENS TO pH: A. pH CAN CHNAGE B. pH WILL RESIST CHANGE TO
A. pH CAN CHNAGE
27
ADDING A STRONG ACID TO A BUFFER CONVERTS A STOICHIOMETRIC AMOUNT OF CJB TO A. WEAK ACID B. WEAK BASE C. STRONG ACID D.STRONG BASE
A. WEAK ACID
28
ADDING A STRONG ACID TO A BUFFER CONVERTS A STOICHIOMETRIC AMOUNT OF WEAK ACID TO A. CJ ACID B. WEAK BASE C. CJ BASE D.STRONG BASE
C. CJ BASE
29
CALCULATING pH AFTER ADDING ACID OR BASE REQUIRES BREAKING THE PROBLEM INTO TWO PARTS:
1. A STOICHIOMETRIC CALCULATION THAT DETERMINES HOW THE ADDITION OF ACID OR BASE CHANGES THE WEAK ACID AND CJB 2. AN EQ CALCULATION OF [H+] USING NEW VALUES OF [HA] AND [A-] IN THE INITIAL ICE BOX VALUES
30
A GOOD BUFFER SHOULD BE ABLE TO NEUTRALIZE GREAT AMOUNTS OF ADDED ACID OR BASE TRUE OR FALSE
FALSE, MODERATE AMOUNTS
31
BUFFERING CAPACITY
THE AMOUNT OF ACID OR BASE A BUFFER CAN NEUTRALIZE WITHOUT CAUSING A LARGE CHANGE IN pH
32
BUFFERING RANGE
THE pH RANGE THE BUFFER CAN BE EFFECTIVE AT
33
THE EFFECTIVENESS OF A BUFFER DEPENDS ON 2 FACTORS
1. RELATIVE AMOUNTS OF ACID AND BASE 2, THE ABSOLUTE CONCENTRATIONS OF ACID AND BASE
34
CONDITIONS FOR AN EFFECTIVE BUFFER: A BUFFER WILL BE MOST EFFECTIVE WHEN:
THE [ACID] AND [BASE] ARE RELATIVELY LARGE. AND WHEN [base]:[acid] = 1
35
CONDITIONS FOR AN EFFECTIVE BUFFER: A BUFFER WILL BE EFFECTIVE WHEN:
THE ACID:BASE RATIO IS: 0.1
36
DETERMINING BUFFERING RANGE: BY SUBSTITUTING THIS EQUATION WE CAN CALCULATE THE MAXIMUM AND MINIMUM pH AT WHICH THE BUFFER WILL BE EFFECTIVE. *HINT* REMEMBER YOU WANT, 0.1
HENDERSON-HASSELBALCH [A-] pH = pKa + log ( ------ ) [HA] LOWEST pH: pH = pKa + log(0.10) HIGHEST pH: pH = pKa + log(10)
37
DETERMINING BUFFERING RANGE: ACCORDING TO, HENDERSON-HASSELBALCH EQUATION, THE EFFECTIVE pH RANGE OF A BUFFER IS A. pKa 1 C. pKa +/- 1
C. pKa +/- 1
38
WHEN CHOOSING AN ACID TO MAKE A BUFFER, CHOOSE ONE WHOSE pKa IS CLOSEST TO THE pH OF THE BUFFER TRUE OR FALSE
TRUE
39
A DILUTED BUFFER CAN NEUTRALIZE MORE ADDED ACID OR BASE THEN A DILUTED BUFFER TRUE OR FALSE
FALSE, A CONCENTRATED BUFFER CAN NEUTRALIZE MORE ADDED ACID OR BASE THEN A DILUTED BUFFER
40
THIS INCREASES THE BUFFERING CAPACITY
INCREASING ABSOLUTE CONCENTRATION BUFFER COMPONENTS
41
AS THE [base]:[acid] RATIO APPROACHES 1, THE ABILITY OF THE BUFFER TO NEUTRALIZE BOTH ADDED ACID AND BASE: A. DECREASES B. IMPROVES
B. IMPROVES
42
A COMMON ION CALCULATION: CALCULATE THE pH OF A BUFFER SOL. THAT IS 0.150 M CH3COOH AND 0.100 M IN NaCH3COO. THE ACID DISSOCIATION CONSTANT FOR CH3COOH IS Ka = 1.8E-5 1) Set up the reaction: 2) Set up the equation, is it an acidic buffer or basic (HINT* C'ION - NO ICE): 3) Plug in and calculate, what is the answer?
1) CH3COOH CH2OO- + H+ 2: ACIDIC BUFFERS SO. HINT* pKa is the -log of A pH = pKa + log [B]/[A] 3) 4.60
43
A COMMON ION CALCULATION: CALCULATE THE pH OF A BUFFER SOLN. THAT IS 0.120 M IN HF AND 0.200 NaF. THE ACID DISSOCIATION CONSTANT FOR HF IS, Ka = 6.8E-4 1) Identify the acid and the base: 2) Set up the equation, is it an acidic buffer or basic (HINT* C'ION - NO ICE): 3) Plug in and calculate, what is the answer?
1) HF is ACID, NaF is Basic. BOTH CREATE THE COMMON ION, F- 2) ACIDIC BUFFERS SO. pH = pKa + log [B]/[A] HINT* pKa is the -log of A 3) 3.39
44
A COMMON ION CALCULATION: HOW MUCH NH4Cl (IN GRAMS) MUST BE ADDED TO 2.0 L OF A 0.10 M NH3 SOLN. TO MAKE A BUFFER THAT HAS A pH = 9.00? THE BASE DISSOCIATION CONSTANT FOR NH3 IS, Kb = 1.8E-5 1) Set up the reaction: What is the equation ? 2) What do you do with pH? * HINT* Find M or [ ]. After what do you do with Kb? 3) Now solve for (g) needed:
1) NH3 + H20 NH4+ + OH- pOH = pKa+ log [A]/[B] 2) 14-9 = 5 -log(Kb) = 4.74 5 - 4.74 = 0.74 = 10E-0.74 = 18 M 3) 2.0 L * 0.18M/ 1L * 53.04g /1 mol = 19.09g
45
CH3COOH, is an Acid. CH3COO-, is a Base. 1) Write the rxn c an acid addition 2) Write the rxn c a base addition
1) HA + CH3 COO- ----> CH3COOH + A- | 2) OH- + CH3COO- ----> CH3COO- + H2O
46
In an acid–base titration, a solution of unknown concentration, slowly added to a solution of known concentration from a burette until the reaction is complete. Is known as a A. Acid B. Base C. Buffer D. Titrant
D. Titrant
47
TITRATION: When the reaction is complete we have reached the endpoint of the titration. Also called the:
Endpoint.
48
A chemical that changes color when the | pH changes
Indicator *May be added to determine the end point, during titration*
49
Titration has reached its equivalence point when the moles of ___ = moles of ___
When the moles of H+ = moles of OH−, the
50
TITRATION CURVES ARE pH VS TITRANT, THE END POINT OF THE CURVE IS CALLED THE EQUIVALENCE POINT OR THE
INFLICTION POINT
51
the Ph of the equivalence point depends on the pH OF THE A. SALT SOLUTION B. ACID VS BASE
A. SALT SOLUTION
52
PRIOR TO THE EQUIVALENCE POINT, THE KNOWN SOLUTION IN THE FLASK IS ________, SO THE pH IS CLOSEST TO IT'S pH. A. LIMITED B. IN EXCESS
B. IN EXCESS
53
EQUIVALENCE POINT OF A NEUTRAL SALT A. pH 7 C.pH = 7
C. pH = 7
54
TITRATING WEAK ACID WITH STRONG BASE: 1) THE INITIAL pH IS DETERMINED USING THE ___ OF THE WEAK ACID 2) THE pH AT THE EQUIVALENCE POINT IS DETERMINED USING THE ___ OF THE CONJUGATE BASE OF THE WEAK ACID A. Ka B. Kb
1- Ka 2-Kb
55
THE pH AFTER EQUIVALENCE POINT IS DOMINATED BY THE A. EXCESS STRONG ACID B. EXCESS STRONG BASE
B. EXCESS STRONG BASE
56
THE BASICITY FROM THE CONJUGATE BASE ANION IS A. NEGLIGIBLE B. NOT NEGLIGIBLE *so small or unimportant as to be not worth considering; insignificant*
A. NEGLIGIBLE *so small or unimportant as to be not worth considering; insignificant*
57
BEFORE THE EQUIVALENCE POINT, THE SOLN. BECOMES A BUFFER. 1) YOU CAN CALCULATE THE mol HAinit. AND mol A-init USING ________. 2) YOU CAN CALCULATE THE pH WITH THE __________ EQUATION USING mol HAinit AND mol A-init
1) RXN STOICHIOMETRY | 2) H-HB EQUATION
58
pH = pK, IS USED TO CALUCLATE
HALF NEUTRILIZATION
59
WHAT IS THE CHANGE IN pH WHEN 5.0 mL OF 4.0 M NaOH IS ADDED TO A 1.00 L BUFFER CONTAINING 0.30 M HF AND 0.30 NaF AT pH = 3.17? (HF, Ka = 6.8E-4) 1) SET UP RXN. *HINT* THIS IS A "CHANGE IN BUFFER, ADD OH-. 2) USE YOUR mL AND M TO CREATE moles OF THE NEW BASE - SET UP ICE 3) SET UP pH EQUATION AND SOLVE FOR DELTA pH HOW MUCH NaOH WOULD NEED TO BE ADDED TO CHANGE pH?
[NaOH] = [OH-] 0.005 L * 4 mol / 1 L = .02 mol OH- OH- + HF ---> F- +H2O 0.020 0.30 0.30 -0.2 -0.2 +0.2 0 0.28 0.32 pH = 3.17 + log [0.32]/[0.28] DELTA pH = 3.23 - 3.17 = .06 MORE THEN 0.30 mol
60
CALCULATE THE CHANGE IN DELTA pH IN A BUFFER: 1) SET UP RXN. *HINT* THIS IS A "CHANGE IN BUFFER, ADD OH-. 2) ICE - SUBTRACT OR ADD BY YOUR LIMITING REAGENT 3) SET UP pH EQUATION AND SOLVE FOR DELTA pH COMPARE TO THE SAME pH OF NaOH ADDED TO 1.0 L OF WATER
A B Na[OH-] + CH3COOH ---> CH3COO- + H2O 0.01 0.10 0.10 -0.01 +0.01 +0.01 0 0.09 0.11 pH = 4.74 + log (0.11/0.09) = 4.83 DELTA pH = 4.83 - 4.74 = 0.09 ``` [OH-] = [NaOH] [OH-] = 0.1 mol / 1 L = 0.1 M pOH = log (0.1 M) pOH = 2.00 -14 = 12.00 pH ```
61
GIVEN HF AND NaOH, SET UP THE RXN
H20 + F- ----> HF + OH-
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