5.1.3 Acids, Bases And Buffers

Question 1

what is a Bronsted-Lowry Acid or Base?

Answer 1

A Bronsted-Lowry acid donates protons (H+)

A Bronsted-Lowry base accepts protons (H+)

Question 2

what are conjugate acid-base pairs?

Answer 2

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

Question 3

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)

Answer 3

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

Question 4

What are monobasic, dibasic and tribasic acids?

Answer 4

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

Question 5

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

Answer 5

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

Question 6

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

Answer 6

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

Question 7

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

Answer 7

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

Question 8

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

Answer 8

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)

Question 9

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

Answer 9

-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

Question 10

how do you calculate pH from H+ ion concentration?

Answer 10

pH = -log[H+(aq)]

Question 11

how do you calculate H+ ion concentration from pH

Answer 11

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

Question 12

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

Answer 12

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

Question 13

How does diluting a solution have an effect on pH?

Answer 13

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

Question 14

what is the relationship between Ka and pKa

Answer 14

pKa = -log Ka

Ka = 10^pKa

Question 15

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?

Answer 15

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

Question 16

what is the weak acid disassociation constant equation?

Answer 16

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

Question 17

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

Answer 17

strong: HA → H+ + A- (non-reversible)

Weak: HA ⇌ H+ + A- (reversible)

Question 18

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

Answer 18

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

Question 19

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

Answer 19

[H+] = √Ka x [HA]

pH = -log[H+]

Question 20

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

Answer 20

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

(H2O + H2O ⇌ H3O+ + OH-)

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

Question 21

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

Answer 21

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

Question 22

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

Answer 22

temperature

Question 23

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

Answer 23

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