5.1.3 Acids, Bases And Buffers Flashcards

1
Q

what is a Bronsted-Lowry Acid or Base?

A

A Bronsted-Lowry acid donates protons (H+)

A Bronsted-Lowry base accepts protons (H+)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

what are conjugate acid-base pairs?

A

A conjugate acid base pair is a substance that contains 2 species that can be interconverted

species that are linked by a transfer of a proton

e.g HCl(aq) ⇌ H+(aq) + Cl-(aq)

[forward reaction] HCl releases a proton to form conjugate base Cl-

[reverse reaction] Cl- accepts a proton to form its conjugate acid HCl

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

how many acid base pairs are there in an acid-base equilibrium?

what are the pairs in this reaction?
HCl(aq) + OH-(aq) ⇌ H2O(l) + Cl-(aq)

A

there are 2 conjugate acid-base pairs so 2 acids (acid 1/2) and 2 bases (base1/2)

HCl = acid 1 bc it donates a proton to OH- so OH- = Base 2 as it accepts a proton

H2O = acid 2 bc it donates a proton to Cl- so Cl- = base 1

THERE CANNOT BE 2 ACIDS/BASES ON THE SAME SIDE, NOR CAN ACID1 AND BASE1 BE ON THE SAME SIDE AND VISEVERSA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

What are monobasic, dibasic and tribasic acids?

A

Monobasic- can donate 1 H+
Dibasic - can donate 2 H+
Tribasic can donate 3 H+

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What volume of 0.1 M HCl (monobasic acid) will neutralise 25cm3 of 0.1M NaOH?

A

25cm3 of HCl because 1 H+ (from HCl) ion neutralises 1 OH- (from NaOH) ion as HCL is monobasic

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What volume of 0.1 M H2SO4 (monobasic acid) will neutralise 25cm3 of 0.1M NaOH?

A

25÷2 because H2SO4 is dibasic and so for each molecule of acid it can dissociate 2 H+ ions and so Neutralise 2 NaOH and so its 12.5cm3 H2SO4

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What volume of 0.1 M H3PO4 (monobasic acid) will neutralise 25cm3 of 0.1M NaOH?

A

25÷3 because H3PO4 is a tribasic acos and so can dissasociate 3 H+ ions into solution so each H3PO4 can neutralise 3 NaOH so = 8.33cm3 H3PO4

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

How do you turn the reactions of acids (e.g with metal oxides, carbonates and Alkalis) into ionc equations?

A

Look at the full equation and cancel out any spectator ions and add in charges and state symbols
e.g
2HCl(aq) + Mg(aq) → MgCl2(aq) + H2(g)

Ionic equation:
2H+(aq) + Mg(s) → Mg2+(aq) + H2(g)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

what does the pH scale tell us in terms of H+ ions?

A
-A low value of [H+] matches a high 
 value of pH
-A high value of [H+] matches a low 
 value of pH
-pH is a logarithmic scale for 
 measuring the [H]+ in solution.
-Each increasing pH value is a 
 magnitude of 10 smaller.

so: pH = -log[H+(aq)]
and: [H+(aq)] = 10^-pH

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

how do you calculate pH from H+ ion concentration?

A

pH = -log[H+(aq)]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

how do you calculate H+ ion concentration from pH

A

[H+(aq)] = 10^-pH

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

How do you calculate the pH of strong acids?

solve this:
HA(aq) → H+(aq) + A-(aq)
[HA]=1.35 x 10^-2moldm-3?

[Hint] Strong acid HA completely dissociates

A

since HA completely dissociates, the conc of HA = conc of H+ since all H+ came from HA
so pH of HA = -log(1.35 x 10^-2 )
= 1.87

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

How does diluting a solution have an effect on pH?

A

changing the concentration
(e.g diluting an acid/base)
can have an effect on the pH as A pH of 1 has 10 times the concentration of H+ ions as a solution of pH 2.

e.g
50cm3 of 0.1M hydrochloric acid is diluted TO A TOTAL VOLUME of 100cm3, what is the pH change?

0.1M has been halved to 0.05M
pH = -log[0.1] = 1
pH = -log[0.05] = 1.3

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

what is the relationship between Ka and pKa

A

pKa = -log Ka

Ka = 10^pKa

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

how do you find the new concentration in a dilution calculation?

e.g. find the pH of HCl if it’s 5cm3 of 0.025moldm-3 and is diluted to 100cm3?

A

do (original volume/new volume) = 5/100
Multiply this by the concentration = 0.05 x 0.025 = 1.25x10-3 moldm-3

then since [H+] = [HA] find the pH =-log 1.25x10-3 moldm-3 = 2.9

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

what is the weak acid disassociation constant equation?

A

Ka= [H+] [A-] / [HA]

17
Q

how do you represent strong and weak acids in an equation?

A

strong: HA → H+ + A- (non-reversible)

Weak: HA ⇌ H+ + A- (reversible)

18
Q

what are the assumptions we make for Ka when calculating something for STRONG acids

A

1) [HA] when dissociated forms equal amounts of both [H+] and [A-] and the dissociation of H2O is negligable

this means that the Ka equation can be simplified to Ka = [H+]^2 / [HA]
as conc of A- = conc of H+

2) the dissociation of weak acids is so small, the change in [HA] is negligible and does not matter

19
Q

what are the equations to find pH and [H+] in relation to Ka?

A

[H+] = √Ka x [HA]

pH = -log[H+]

20
Q

what is the expression for the ionic product of water? (Kw)

A

Ka = [H+] [OH-] / [H2O]

(H2O + H2O ⇌ H3O+ + OH-)

So if any question mentions OH- ions, it i=uses Kw

21
Q

what is the value for the constant Kw? (ionic product of water)

A

1.00x10^-14 mol^2dm^-6 at 298 K

22
Q

what is the only thing that can affect the values of Kw?

A

temperature

23
Q

what are the problems with the Ka calculation method and assumptions?

A

1) when calculating the values for Ka we assume the the concentrations of H+ and A- are the same, but for acids with a pH less than 6, the disassociation of H2O becomes more significant, so [H+] is no longer = to [A-]

so [H+] = [A-] assumption does not work for very weak acids or dilute solutions

2) the second assumption assumes