Acid-base Equilibria

Question 1

Brønsted-Lowry acid

Answer 1

Proton donor
H+ combines with H2O to form H3O+

Question 2

Brønsted-Lowry base

Answer 2

Proton acceptor
Forms water

Question 3

Strong acid

Answer 3

Dissociate almost completely in water

Question 4

Weak acid

Answer 4

Dissociate only slightly in water
Equilibrium is set up where equilibrium lies to the left

Question 5

Conjugate pairs

Answer 5

Species that are linked by the transfer of a proton

Question 6

Conjugate base

Answer 6

Species that has lost a proton

Question 7

Conjugate acid

Answer 7

Species that has gained a proton

Question 8

Neutral solutions

Answer 8

[H+]=[OH-]
All of each ion reacts to form water

Question 9

Standard enthalpy change of neutralisation

Answer 9

Enthalpy change when solutions of an acid and a base react together under standard conditions to produce one mole of water

Question 10

Define pH

Answer 10

Measure of the hydrogen ion concentration

Question 11

Calculating pH

Answer 11

-log10[H+]
(Don’t type the 10 in calculator)

Question 12

Calculating [H]+ from pH

Answer 12

10^-pH

Question 13

Weak base

Answer 13

Partially react with water to form OH-

Question 14

Calculating pH of a strong acid

Answer 14

As they fully dissociate, [acid]=[H+]

Question 15

Calculating pH of a weak acid

Answer 15

Don’t fully dissociate so need Ka and [acid]
Assume that the [H+]=[X-]
Ka= [H+]^2 / [acid]

Question 16

Monoprotic

Answer 16

Each mole of acid produces one mole of H+

Question 17

Poly protic

Answer 17

Each mole of acid produces more than one mole of H+

Question 18

Assumption from textbook for calculating pH of weak acids

Answer 18

Dissociation of acid is much stronger than that of water (all H+ come from the acid)
[acid at start]=[acid at equilibrium]
[base]=[H+]

Question 19

Calculating pH of a strong base

Answer 19

Fully dissociate to release OH-
[OH-]=[base]
Need Kw and [OH-] to calculate [H+] then pH

Question 20

Ionic product of water

Answer 20

H2O—-> H+ + OH-
<—-
So Kc expression applies
Water only dissociates a tiny amount so equilibrium lies to left so consider [H2O] to have a constant value. So Kc multiplied by [H2O] makes a constant, Kw
Pure water, [H+]={OH-] so [H+]^2

Question 21

Define pKa and pKw

Answer 21

PKw=-log10Kw
Kw=10^-pKa
PKa=-log10Ka
Ka=10^-pKa

Question 22

Calibrating a pH metre

Answer 22

Place the bulb of the pH meter into deionised water and allow the reading to settle and adjust the reading so that it says 7
Repeat for a standard solution of pH4 and 10 (rinsing the meter with deionised water between each reading)

Question 23

PH of salts of strong acids and strong bases

Answer 23

Neutral

Question 24

Using masses and pH to calculate Ka for weak acids

Answer 24

Calculate moles of the acid
Calculate concentration of the acid solution
Use pH to work out concentration of H+ at equilibrium
Ka=[H+]^2/[HA]
Remember [HA] start=[HA]equilibrium

Question 25

Diluting a strong acid

Answer 25

Diluting by a factor of: 10: increases pH by 1

Question 26

Diluting a weak acid by a factor of 10 Formula for pH of a weak acid

Answer 26

increases pH by 0.5 PH=-log10 root(Ka[acid])

Question 27

Titration curves

Answer 27

Plot pH against amount of base added Equivalence line has the equivalence point, where [H+]=[OH-], neutralisation. Small addition of base causes huge change in pH Initial pH depends on strength of the acid Final pH depends on strength of the base

Question 28

Using titration curves to find pKa

Answer 28

Find the pH at half the equivalence point

Question 29

Practical

Answer 29

Rinse burette with distilled water then with the solution filling it Calibrate the pH probe and rinse it with distilled water Perform rough titre and record pH in increments of volume of solution in burette to find approximate equivalence point (PH when just pink if using indicator then add more of the acid and record that pH) Repeat, decreasing size of increments as getting close to rough titre Graph of pH against volume of burette solution added

Question 30

Kw at 298K

Answer 30

1x10^-14 mol^2dm^-6

Question 31

Amphoteric

Answer 31

Acts as a base and an acid Water