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Chemistry > 1.10 acids and bases > Flashcards

1.10 acids and bases Flashcards

1
Q

what is a bronsted-lowry acid?

A

proton donors- release H+ when mixed with water

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2
Q

what does bronsted lowry theory say about H+ ions?
equation?

A

theyre never by themselves in water- theyre always combined with H2O to form hydroxonium ions (H3O)
HA(aq) + H2O (l) -> H3O+ (aq) + A-(aq)

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3
Q

what is a bronsted- lowry base?

A

a proton acceptor

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4
Q

what do bronsted- lowry bases do in solution?
equation?

A

grab hydrogen ions from water molecules
B(aq) + H2O(l) -> BH+(aq) + OH-(aq)

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5
Q

what does it mean that acids and bases can dissociate in water?

A

they break up into positively and negatively charged ions

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6
Q

what is a strong acid/ base?

A

dissociates almost completely in water- nearly all H+ ions will be released

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7
Q

what is a weak acid/ base?

A

dissociate only very slightly in water- only a small number of H+ ions are formed.
an equilibrium is set up which lies well over to the left

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8
Q

what happens when an acid is added to water?
equation?

A

the water acts as a base and accepts the proton
HA(aq) + H2O <-> H3O+(aq) + A-(aq)

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9
Q

what is the equation for water dissociating?
simplified?

A

H2O(l) + H2O(l) <-> H3O+(aq) + OH-(aq)
H2O (l) <-> H+ (aq) + OH-(aq)

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10
Q

where does the equilibrium lie for the dissociation of water?

A

well to the left as it only dissociates a small amount

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11
Q

what is Kw?

A

eq for dissociation of water lies so left that there is so much water compared to ions that conc of water is seen as constant

so, multiply expression for Kc of dissociation of water by [H2O], you get a constant- the ionic product of water (Kw)

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12
Q

Kw=

A

[H+][OH-]

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13
Q

what is the value of Kw?

A

1.00 x 10^-14 mol2dm-6

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14
Q

when does Kw change?

A

when temp changes

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15
Q

what happens to Kw when dealing with pure water?

A

in pure water, theres always 1 H+ for 1 OH- so you can say that Kw= [H+]^2

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16
Q

what is the equation for calculating pH?

A

pH= -log10[H+]

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17
Q

what is the equation for calculating hydrogen ion concentration?

A

[H+] = 10^-pH

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18
Q

what does monoprotic mean?

A

each molecule in an acid will release one proton when it dissociates eg HCl and HNO3

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19
Q

how can you work out pH of a monoprotic acid?

A

They release 1 mol of H+ ions per mol of acid so H+ conc is same as acid conc so you can work out pH

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20
Q

what is a diprotic acid?

A

each molecule of acid will release 2 protons when it dissociates eg H2SO4

produce 2 mol of H+ per mol of acid so H+ conc is twice conc of the acid

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21
Q

how many decimal places should you write pH to?

A

2

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22
Q

what is the expression for Kw?

A

Kw= [H+][OH-]

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23
Q

in strong monoprotic bases, why is [OH-] same and [base]

A

because they fully dissociate in water so for every mole of base, they donate one mole of OH- ions

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24
Q

how do you work out pH of strong bases?

A

find values of Kw and [OH-] (probably from question)

rearrange the equation for [H+]
([H+]= Kw/ [OH-])

substitute this into pH equation (pH=log[H+])

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25
what is the acid dissociation constant?
weak acids only partly dissociate in water so [H+] isnt same as acid concentration so Ka is used not Kw
26
what is the expression for Ka? Units?
Ka= [H+][A-]/[HA] moldm-3
27
how do you derive Ka?
HA<-> H+ + A- only partially dissociates so can assume [HA] start is same as [HA] equilibrium
28
how can you simplify Ka equation? why?
Ka= [H+]^2 / [HA] when dealing with weak acids, you can assume all the H+ ions come from acid so [H+]= [A-]
29
how do you find the pH of weak acids?
write an expression for Ka for the weak acid rearrange to find [H+]^2 square root this use -log[H+]
30
how do you find the concentration of weak acids?
find [H+] using 10^-pH write expression for Ka rearrange to find concentration substitute in values and solve
31
how do you find Ka of weak acids
if you know conc or pH, you can use them to find the Ka of weak acid find [H+] using 10^-pH substitute values into equation for Ka
32
what is pKa?
Ka varies a lot from one acid to next but this can make the numbers hard to manage so pKa is easier
33
how do you work out pKa?
pKa= -log10(Ka) Ka=10^-pKa
34
pH curves practical
35
what does a pH curve for strong acid and strong base look like?
pH starts around1 as theres an excess of acid finishes arounf 13 when you have an excess of strong base
36
what does a pH curve for strong acid and weak base look like?
pH starts around 1 as theres an excess of strong acid finishes around 9 when you have an excess of weak base
37
what does a pH curve for weak acid and strong base look like?
pH starts around 5 as theres an excess of weak acid and finishes around 13 when theres an excess of strong base
38
what does a pH curve for weak acid and weak base look like?
pH starts around 5 as theres an excess of weak acid finishes around 9 as theres an excess of weak base
39
what is the equivalence/ end point of a pH curve?
the mid-point of the vertical section atp, a tiny amount of base causes a sudden, big change in pH
40
why is it easier to use a pH meter to find the end point of a weak acid/ weak base titration?
you dont get such a sharp change- colour changes gradually so hard to see exact end point no vertical section on the pH curve
41
why doesnt it matter that phenolphthalein and methyl orange dont change colour at exactly pH7?
the range that it changes in is in the vertical part of the pH curve so despite drastic changes in pH, change in vol is only very small
42
methyl-orange and phenolphthalein colours and pH
methyl-orange low= red, high= yellow, range= 3.1- 4.4 phenolphthalein low= colourless, high= pink, range= 8.3 - 10
43
indicator for strong acid/ strong base titration
either rapid pH change over range for both indicators
44
indicator for strong acid/ weak base titration
methyl orange pH changes rapidly across range for methyl orange but not phenolphthalein
45
indicator for weak acid/ strong base titration
phenolphthalein pH changes rapidly over phenolphthalein's range but not methyl orange's
46
indicator for weak acid/ weak base titration
pH meter no sharp pH change so no indicator will work
47
how to work out conc of diprotic acids using titration calculations
reaction happens in 2 stages as 2 protons are removed separately so you get a pH curve with 2 equivalence points calculate conc same as for monoprotic but need twice as many moles of base as moles of acid
48
how to calculate conc from pH curve?
write balanced eq use graph to see vol of reactant (draw line down from equivalence point) see what else youve been given work out as usual
49
what is a buffer?
a solution that resists change in pH when small amounts of acid or alkali are added doesnt stop pH changing completely but makes the changes very slight. only work for small amounts of acid/base
50
what are acidic buffers?
buffers with a pH less than 7 mixture of weak acid with one of its salts can resist change in pH when either an acid or base is added to the solution
51
how do acidic buffers work?
eg ethanoic acid and CH3COO-Na+ ethanoic acid is weak so only slightly dissociates but salt fully dissociates so solution has lots of undissociated ethanoic acid and ethanoate ions (from salt). when you alter H+ or OH- conc, eqm position moves to counteract
52
how do acidic buffers resist an acid?
2 eqm equations coexist in same beaker: CH3COOH <-> CH3COO- + H+ eqm left CH3COO-Na+ -> CH3COO- + Na+ eqm right - add H+, it reacts with the CH3COO- - high conc of these from the salt but adding H+ means more CH3COOH so eqm shifts left
53
how do acidic buffers resist a base?
2 eqm equations coexist in same beaker: CH3COOH <-> CH3COO- + H+ eqm left CH3COO-Na+ -> CH3COO- + Na+ eqm right - OH- reacts w H+ ions - low conc of these but can be reproduced from high conc of CH3COOH to counteract change (le chateliers principle) - eqm shifts right to replace H+ ions
54
what is a basic buffer?
a buffer with a pH that is greater than 7 contains mixture of weak base and one of its salts can resist changes in pH when acid or base is added
55
how do basic buffers work?
eq NH3 and NH4Cl salt fully dissociates and some of the NH3 molecules also react w water so solution has lots of NH4+ and ammonia. eqm changes to counteract pH
56
how do basic buffers resist acids?
2 eqm equations coexist in same beaker: NH3 + H2O <-> NH4+ +OH- eqm left NH4+Cl- -> NH4+ + Cl- eqm right - H+ reacts w OH- which has a low conc but is reproduced from NH3 + H2O (le chataliers principle) - eqm shifts right
57
how do basic buffers resist bases?
2 eqm equations coexist in same beaker: NH3 + H2O <-> NH4+ +OH- eqm left NH4+Cl- -> NH4+ + Cl- eqm right - OH- reacts w NH4+ which is in high conc - More NH3 + H2O produced so eqm shifts left
58
how do buffers resist changes in pH when diluted by water?
small amount of water added, water slightly dissociates so extra H+ and OH- ions push eqm same amount in both directions, leaving it unchanged
59
what are the applications of buffers?
- alkaline conditions make surface of hair rougher so buffers added to shampoos to keep pH around 5.5 to keep hair smooth - biological washing powders- keep pH at right level for enzymes to work best - in our body- make sure tissues are kept at right pH eg blood has to stay very near 7.4
60
how to calculate the pH of buffers using known concentrations
1. write expression for Ka of weak acid 2. rearrange to give expression for [H+] 3. sub in Ka and concs of acid and salt (when using Ka, use eqm concs) 4. solve to find [H+] 5. sub this into pH eq and solve
61
another way of making acidic buffers?
can mix weak acid and add small amount of alkali so some acid is neutralised to make a salt and some left unneutralised.
62
how to calculate pH of acidic buffer made from acid and alkali
1. write eq for neutralisation reaction 2. calculate mol of acid and base at start using vols and concs given 3. use molar ratios in eq to work out mol of acid and salt left at end 4. calculate conc of acid and salt in buffer solution- this is vol of acid and base together 5. calculate pH as before (using Ka)
63
what is half neutralisation?
half way between 0 and equivalence point can use to calculate pKa of weak acid by taking pH at this point. [HA]=[A-] so they cancel in the Ka expression so ka=[H+] -log(ka)=-log(H+) so pka=pH

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