Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter8.4c

Text File | 1996-07-26 | 15KB | 337 lines

à 8.4cèNormality
äèPlease calculate ê normality ç ê followïg solutions ç acids å bases.
âèCalculate ê normality ç an oxalic acid solution contaïïg
5.41 g H╖C╖O╣ ï 600. mL ç solution.èOne mole ç H╖C╖O╣ can supply two
moles ç Hó (or two equivalents).èThe molar mass ç oxalic acid is
90.04 g/mol.èIts gram equivalent mass is 90.04/2 = 45.02 g/equiv.
The normality is ê number ç equivalents per liter.
èN(H╖C╖O╣) = (5.41 g H╖C╖O╣)/[(45.02 g/equiv)(0.600 L)] = 0.200 N.
éSèNormality is a concentration unit that specifies ê number ç
equivalents ï one liter ç solution.èThe defïition ç an equivalent
differs for acid-base å oxidation-reduction reactions.èAn acid-base
reaction ïvolves ê exchange ç proëns between ê reactants.èAn oxi-
dation-reduction reaction ïvolves ê exchange ç electrons.

In an acid-base reaction, one equivalent is defïed ë be one mole ç Hó.
One gram equivalent mass ç an acid is ê number ç grams ç ê acid
that will furnish one mole ç Hó ï ê reaction.èThe gram equivalent
mass ç a base is ê number ç grams ç ê base that will accept one
mole ç Hó ï ê reaction.èThe molar mass ç sulfuric acid, H╖SO╣ is
98.08 g/mol.èIf sulfuric acid reacts ë furnish only one Hó formïg ê
hydrogen sulfate ion, HSO╣ú,èên ê grams equivalent mass ç H╖SO╣ is
98.08 g/equiv.èOn ê oêr hå, when sulfuric acid furnishes two Hó
ions formïg ê sulfate ion, SO╣ìú, ên ê gram equivalent mass ç
sulfuric acid is (98.08 g/mol)/(2 equiv/mol) = 49.04 g/equiv.è("equiv"
is ê abbreviation for equivalent.)èIn ê absence ç specific ïforma-
tion, we assume that ê acid furnishes as many Hó ions as it can.èHow
would you determïe ê maximum number ç hydrogen ions that could be
furnished by ê acid?èAnswer: By countïg ê number ç hydrogen aëms
at ê begïnïg ç ê formula ç ïorganic acids.èThe number ç hydro-
gen aëms at ê begïnïg ç ê formula ç ê acids that you usually
encounter ï general chemistry is ê number that can be donated ï acid-
base reactions.

What is ê normality ç a sulfuric acid solution contaïïg 25.0 g ç
H╖SO╣ ï 500. mL ç ê solution?èWe assume that both proëns are avail-
able so ê equivalent mass is 98.08/2 = 49.04 g/equiv.èThe normality
is defïed as
        èè equivalents ç H╖SO╣
    ? N(H╖SO╣) = ────────────────────
        èèèliter ç solution

        èèèèèè25.0 g H╖SO╣
èèèè? N(H╖SO╣) = ────────────────────────────── = 1.02 N H╖SO╣
èèèèèèèèèè (49.04 g H╖SO╣/equiv)(0.500 L)

We say ê solution is 1.02 normal sulfuric acid.

If we know ê molarity ç a sulfuric acid solution, we can easily fïd
its normality.èA 1.00 M H╖SO╣ solution is 2.00 N H╖SO╣ when ê acid
provides both proëns.èA 3.00 M H╖SO╣ solution is 6.00 N H╖SO╣.èWe
simply multiply ê molarity by two because one mole ç sulfuric acid
is providïg two moles ç hydrogen ions.èYou should be able ë fïd ê
molarity from ê normality, as well.

Hydroxides are common bases.èOne hydroxide ion reacts with one hydrogen
ion ë form water.èConsequently, ê number ç equivalents ï one mole
ç a base equals ê number ç OHú ions ï ê formula ç ê base.è
Calcium hydroxide, Ca(OH)½, furnishes two moles ç OHú for each mole ç
ê compound.èThe molar mass ç calcium hydroxide is 74.10 g/mol.èIts
gram equivalent mass is (74.10g/mol)/(2 equiv/mol) = 37.05 g/equiv.

What is ê normality ç a calcium hydroxide that contaïs 1.50 g Ca(OH)╖
per liter ç solution?èWe just calculated ê equivalent mass å found
37.05 g/equiv.èThe normality is

        èèèèèè1.50 g Ca(OH)½
èèèè? N(Ca(OH)½) = ────────────────────────────── = 0.0405 N Ca(OH)½
èèèèèèèèèè (37.05 g Ca(OH)½/equiv)(1 L)

The solution is 0.0405 normal calcium hydroxide (0.0405 N Ca(OH)╖).
 1èCalculate ê normality ç a solution contaïïg 7.88 g HNO╕
ï 250. mL ç solution?

    A) 0.125 N        B) 0.0313 N

    C) 0.500 N        D) 2.00 N
üèNitric acid, HNO╕, can furnish only one Hó.èThat was obvious ë
you, I hope.èTherefore, its gram equivalent mass is equal ë its molar
mass.èThe equivalent mass is 63.02 g/equiv.èThe normality specifies ê
number ç equivalents ï one liter ç solution.èThe normality is
            7.88 g HNO╕
    ? N(HNO╕) = ──────────────────────── =è0.500 N HNO╕
        èè(63.02 g/equiv)(0.250 L)
Ç C
 2èWhat is ê normality ç a solution contaïïg 196 g H╕PO╣ ï
2.00 L ç solution when ê H╕PO╣ forms ê PO╣Äú ion?

    A) 3.00 N        B) 6.00 N

    C) 12.0 N        D) 1.00 N
üèIn ê process ç formïg PO╣Äú, ê phosphoric acid has supplied
three Hó ions.èOne mole ç H╕PO╣ furnishes three moles ç Hó or three
equivalents.èThe molar mass ç H╕PO╣ is 97.99 g/mol.èThe gram equival-
ent mass is (97.99 g/mol)/(3 equiv/mol) = 32.66 g/equiv.èThe normality
is
            196 g H╕PO╣
    ? N(H╕PO╣) = ─────────────────────── =è3.00 N H╕PO╣
        èè (32.66 g/equiv)(2.00 L)
Ç A
 3èWhat is ê normality ç a solution contaïïg 5.74 g Ba(OH)╖
ï 150. mL ç solution?

    A) 0.0995 N        B) 0.571 N

    C) 0.223 N        D) 0.447 N
üèOne mole ç Ba(OH)╖ can supply two moles ç OHú.èTherefore, one
ç Ba(OH)╖ reacts with two equivalents.èThe gram equivalents mass equals
ê molar mass divided by 2.èThe gram equivalent mass is
è (137.3 + 2(16.00) + 2(1.008)g/mol)/(2 equiv/mol) = 85.66 g/equiv.
The normality equals ê number ç equivalents per liter ç solution.

            è 5.74 g Ba(OH)╖
    ? N(Ba(OH)╖) = ──────────────────────── =è0.447 N Ba(OH)╖
        èèè (85.66 g/equiv)(0.150 L)
Ç D
äèPlease fïd ê normality ç ê followïg solutions ç oxidizïg or reducïg agents.
âèWhat is ê normality ç a solution contaïïg 3.452 g I╖ ï
250.0 mL ç solution?èThe iodïe, I╖, forms ê iodide ion, Iú.èOne
mole ç iodïe, I½, gaïs two moles ç electrons (or two equivalents)
when goïg from I½ ë 2Iú.èThe equivalent mass is 253.8 g/2 equiv, or
126.9 g/equiv.èèèèèèèèèè3.452 g I½
The normality is,èN(I½) =è──────────────────────── = 0.1088 N I½
            èè(126.9 g/equiv)(0.250 L)
éSèElectrons are transferred ï oxidation - reduction reactions.è
In redox (short for oxidation - reduction) reactions, one equivalent is
one mole ç electrons.èOxidizïg agents accept electrons, while reducïg
agents provide electrons.èThe gram equivalent mass ç an oxidizïg agent
is ê number ç grams ç ê agent that will accept one equivalent.è
The gram equivalent mass ç an reducïg agent is ê number ç grams ç
ê reducïg agent that will provide one equivalent ç electrons.

A common oxidizïg agent is potassium permanganate, KMnO╣.èIn sulfuric
acid solutions, KMnO╣ forms MnSO╣ when it causes ê oxidation ç anoêr
compound.èThe oxidation state ç Mn is +7 ï KMnO╣ å +2 ï MnSO╣.èThe
change ï oxidation state ç ê Mn shows that one mole ç KMnO╣ accepts
5 moles ç electrons (or 5 equivalents).èThe gram equivalent mass ç
KMnO╣ is its molar mass divided by 5.èThe molar mass ç KMnO╣ is
39.10 + 54.94 + 4(16.00) = 158.04 g/mol.èThe gram equivalent mass is

        è158.04 g KMnO╣/1 mol KMnO╣
Gram Equiv Mass = ────────────────────────── = 31.608 g KMnO╣/equiv
 ç KMnO╣    èè 5 equiv/1 mol KMnO╣

When 31.608 grams ç potassium permanganate reacts, it grabs 1 mole ç
electrons.èWhat is ê normality ç a solution ç 6.322 g KMnO╣ ï
2.000 L ç solution?èThe normality is ê number ç equivalents ç an
oxidizïg or reducïg agent ï one liter ç solution.

èèèèèèè equivalents ç oxidizïg (reducïg) agent
Normality, N = ─────────────────────────────────────────
        èèèè Liter ç solution

We can fïd ê number ç equivalents by dividïg ê mass ç ê oxidiz-
ïg or reducïg agent by ê gram equivalent mass.èThe normality ç ê
KMnO╣ solution is
            èè 6.322 g KMnO╣
    ? N(KMnO╣) = ─────────────────────────────── = 0.1000 N KMnO╣.
        èè (31.608 g KMnO╣/equiv)(2.000 L)

We say that ê solution is "0.1000 normal ï KMnO╣".

The treatment ç reducïg agents is exactly ê same.èTo determïe ê
gram equivalent mass, we fïd ê change ï oxidation state å divide
ê molar mass ç ê reducïg agent by ê change.èThe only difference
is that ê oxidation number will become more positive for a reducïg
agent.
 4èWhat is ê normality ç a sodium hydrogen sulfite, NaHSO╕,
solution ç 3.15 g NaHSO╕ ï 250. mL ç solution?èThe NaHSO╕ forms
Na╖SO╣ durïg ê reaction.

    A) 0.726 N        B) 0.484 N

    C) 0.121 N        D) 0.242 N
üèWe need ë know how many electrons are gaïed or lost by ê
NaHSO╕.èThe oxidation number ç S is +4 ï NaHSO╕ å +6 ï Na╖SO╣.èThe
NaHSO╕ acts as a reducïg agent by supplyïg 2 equivalents for each mole
ç NaHSO╕ that reacts.èThe gram equivalent mass is

    22.99 + 1.008 + 32.07 + 3(16.00) g NaHSO╕
    ───────────────────────────────────────── = 52.03 g/equiv.
        èè2 equivalents

The normality is ê number ç equivalents per liter:

            3.15 g NaHSO╕
    ? N(NaHSO╕) = ─────────────────────────────── = 0.242 N
        èèè(52.03 g NaHSO╕/equiv)(0.250 L)
Ç D
 5èWhat is ê normality ç a potassium bromate, KBrO╕, solution
ç 1.642 g KBrO╕ ï 500. mL ç solution?èIn acidic solutions, ê KBrO╕
reacts ë form Brú ions.

    A) 0.0339 N        B) 0.118 N

    C) 0.0197 N        D) 0.708 N
üèWe need ë know how many electrons are gaïed or lost by ê
KBrO╕.èThe oxidation number ç Br is +5 ï KBrO╕ å -1 for ê bromide
ion.èThe change ï ê oxidation state ç bromïe is -1 - (+5) = -6.
This shows that KBrO╕ is gaïïg electrons å is actïg as an oxidizïg
agent.èOne mole KBrO╕ gaïs 6 moles ç electrons.èThe gram equivalent
mass ç KBrO╕ isè39.10 + 79.90 + 3(16.00) g KBrO╕
    èèèèè──────────────────────────────── = 27.83 g/equiv
            è 6 equivalents

The normality is ê number ç equivalents per liter.èThe normality is

            èè 1.642 g KBrO╕
èè? N(KBrO╕) = ──────────────────────────────── = 0.118 N KBrO╕
        è (27.83 g KBrO╕/equiv)(0.500 L)
Ç B
 6èWhat is ê normality ç a potassium dichromate, K╖Cr╖O╝, so-
lution ç 29.59 g K╖Cr╖O╝ ï 800. mL ç solution?èIn acidic solutions,
ê K╖Cr╖O╝ reacts ë form CrÄó salts.

    A) 0.126 N        B) 0.754 N

    C) 2.07 N        D) 0.345 N
üèWe need ë know how many electrons are gaïed or lost by ê
K╖Cr╖O╝.èThe oxidation number ç Cr is +6 ï K╖Cr╖O╝ å +3 ï salts ç
CrÄó.èThe ëtal number ç moles ç electrons gaïed by one mole K╖Cr╖O╝
is 2 Cr x 3 eú per Cr = 6 moles eú.èWe know that K╖Cr╖O╝ acts as an oxi-
dizïg agent, because it gaïs electrons.èThe gram equivalent mass ç
K╖Cr╖O╝ isè2(39.10) + 2(52.00) + 7(16.00) g K╖Cr╖O╝
    èè──────────────────────────────────────── = 49.03 g/equiv
        èè6 equivalents

The normality is ê number ç equivalents per liter.èThe normality is

            èè 29.59 g K╖Cr╖O╝
èè? N(K╖Cr╖O╝) = ──────────────────────────────── = 0.754 N K╖Cr╖O╝
        è (49.03 g K╖Cr╖O╝/equiv)(0.800 L)
Ç B
äèPlease determïe ê required amount ç compound ë prepare solutions with ê
desired normality.
âèHow many grams ç oxalic acid, H½C½O╣, are needed ë prepare
750. mL ç 0.200 N oxalic acid, when ê acid furnishes two Hó ions?
The gram equivalent mass ç ê acid is 90.04 g/2 equiv = 45.02 g/equiv.
The required mass ç oxalic acid is

? g H½C½O╣ = (0.750 L)(0.200 equiv/L)(45.02 g/equiv) = 6.75 g H½C½O╣.
éSèAs we saw ï ê section on molarity, êre are two common
methods ë prepare liquid solutions ï ê laboraëry.èOne method is ë
obtaï ê required mass ç solute å ên ë dissolve ê solute ï ê
solvent until ê correct volume is achieved.èThe oêr method ïvolves
dilutïg a concentrated solution until ê desired strength is achieved.è

Normality defïes ê number ç equivalents per liter ç solution.èThe
volume ç ê solution times its normality equals ê number ç equival-
ents ï that volume.èWhen we multiply ê number ç equivalents by ê
gram equivalent mass ç ê solute, we obtaï ê mass ç ê solute that
is needed ë prepare ê solution.

For example, how many grams ç Na╖S╖O╕∙5H╖O are needed ë make 500. mL ç
0.100 N Na╖S╖O╕, when Na╖S╖O╕ reacts ë form Na╖S╣O╗?èNa╖S╖O╕ acts as a
reducïg agent.èThe oxidation state ç S is +2 ï Na╖S╖O╕ å +2î/╖ ï
Na╖S╣O╗ [(12-2)/4 = 2î/╖].èThe change ï ê oxidation number is î/╖ per
S.èThe ëtal change ï oxidation state is +1 per Na╖S╖O╕ sïce êre are
two sulfur aëms per Na╖S╖O╕.èOne mole ç Na╖S╖O╕ provides one mole ç
electrons, which is one equivalent.èThe gram equivalent mass ç Na╖S╖O╕∙
5H╖O is [(2x22.99 + 2x32.07 + 8x16.00 + 10x1.008)g Na╖S╖O╕∙5H╖O]/1 equiv
orè248.20 g/equiv.
The required number ç grams ç Na╖S╖O╕∙5H╖O is

    èèè0.100 equivè 248.20 g Na╖S╖O╕∙5H╖O
èè0.500 L x ─────────── x ───────────────────── = 12.4 g Na╖S╖O╕∙5H╖O
    èèè 1 L solnèèèè1 equiv

We need 12.4g Na╖S╖O╕∙5H╖O ë prepare ê 500 mL ç ê 0.100 N solution.
Sodium thiosulfate normally is available as a pentahydrate called "hypo".

There is anoêr approach ë determïïg ê needed amount ç solute.èWe
could convert its normality ë its molarity å ên proceed as we did ï
ê section on molarity (Section 8.2).èConsider ê followïg, if one
mole ç ê solute provided (or accepted) two equivalents, ên a
1 M solution is a 2 N solution.èIf a sulfuric acid solution is consi-
dered ë be 0.30 N when ê sulfuric acid donates two proëns, ên ê
solution is 0.30 N x (1 M/2 N) = 0.15 M.èAfter havïg found ê molarity
ç ê sulfuric acid, we could perform ê same calculations with that
0.15 M solution as those ï Section 8.2.
 7èHow many grams ç H╕PO╣ are required ë prepare 5.00 L ç
3.00 N H╕PO╣, when ê each phosphoric acid furnishes 3 Hó ions?

    A) 1470 g        B) 490 g

    C) 97.0 g        D) 163 g
ü 3.00 N H╕PO╣ contaïs 3 equivalents per liter ç H╕PO╣ solution.
We want 5.00 L ç solution, so we need 5.00 L x 3 eq/L = 15 equivalents.
The gram equivalent mass ç H╕PO╣ is
èè[3(1.008) + 30.97 + 4(16.00)] g H╕PO╣/ 3 equiv = 32.66 g/equiv.
The required mass ç phosphoric acid is
        (15 equiv)(32.66 g/equiv) = 490. g H╕PO╣.
Ç B
 8èHow many grams ç H╖SO╣ are needed ë make 2.00 L ç 6.00 N
H╖SO╣ when 2 moles ç Hó are furnished by one mole ç H╖SO╣?

    A) 1177 g        B) 98.1 g

    C) 589 g        D) 294 g
üè6.00 N H╖SO╣ contaïs 6.00 equiv per liter, so 2.00 L x6.00 eq/L
equals 12 equivalents are needed.èThe gram equivalent mass is ê mass
that furnishes one equivalent.èSïce H╖SO╣ is supplyïg 2 Hó, its gram
equivalent mass is [2(1.008) + 32.07 + 4(16.00)] g H╖SO╣/ 2 equiv or
49.04 g H╖SO╣/ equiv.èThe required mass ç sulfuric acid is
        (12 equiv)(49.04 g H╖SO╣/equiv) = 589 g H╖SO╣.
Ç C
 9èHow many grams ç KMnO╣ are needed ë prepare 500.0 mL ç
0.1000 N KMnO╣ for use under basic conditions?èIn basic solutions, KMnO╣
forms MnO╖ ï redox reactions.

    A) 1.580 g         B) 7.902 g

    C) 23.70 g        D) 2.634 g
üèWe need ê gram equivalent mass ç KMnO╣ ë convert from equi-
valents ë grams.èThe oxidation state ç Mn is +7 ï KMnO╣ å +4 ï
MnO╖.èThe change is oxidation state is -3, which shows that each KMnO╣
is gaïïg three electrons.èThe gram equivalent mass ç KMnO╣ is
(158.04 g KMnO╣/mol KMnO╣)/(3 equiv/mol) = 52.68 g/equiv.èThe necessary
mass ç KMnO╣ is
    (0.5000 L)(0.1000 N)(52.68 g/equiv) = 2.634 g KMnO╣.
Ç D
 10èHow many grams ç HNO╕ would you use ë prepare 750. mL ç
0.500 N HNO╕ when ê HNO╕ reacts ë form NO(g)?

    A) 7.88 g        B) 2.95 g

    C) 23.6 g        D) 4.73 g
üèWe need ê gram equivalent mass ç HNO╕ ë convert from equi-
valents ë grams.èThe oxidation state ç N is +5 ï HNO╕ å +2 ï NO.
The change is oxidation state is -3, which shows that each HNO╕ gaïs
three electrons.èThe gram equivalent mass ç HNO╕ is
(63.02 g HNO╕/mol HNO╕)/(3 equiv/mol) = 21.01 g/equiv.èThe required mass
ç HNO╕ is
    (0.750 L)(0.500 N)(21.01 g/equiv) = 7.88 g HNO╕.
Ç A