12: Acids and Bases Flashcards

Question

What is pH?

Answer 1

pH = - log₁₀[H⁺] Normally goes from 0 to 14 7 is regarded as neutral

Answer 2

0 - very acidic 14 - very basic/alkaline

Answer 3

-log₁₀

Answer 4

One mole of acid produces one mole of H⁺ ions

Answer 5

[HA] = [H⁺] Fully dissociates and produces one H⁺per mole of acid

Answer 6

[H⁺] = 10^-pH

Answer 7

An acid that releases more than one proton per molecule upon dissociation

Answer 8

2 moles of H⁺ions are released for every mole of acid which dissociates

Answer 9

Acid dissociation constant Applies to a particular acid at a specific temp regardless of concentration

Answer 10

Ka = [H⁺][A^-] / [HA]

Answer 11

Assume [HA]_inital ≈ [HA]_equilibrium Assume acid dissociates more than water so all protons are from the acid, meaning [H⁺] ≈ [A^-] Therefore Ka = [H⁺]²/[HA]_inital Then use to find [H⁺] and pH

Answer 12

Strong acids dissociate more in solution so the difference is significant

Answer 13

Acid - donating a proton or Base - accepting a proton Always OH^- and H₃O⁺ in water

Answer 14

H₂O + H₂O ⇔ H₃O⁺ + OH^- Simplified to: H₂O ⇔ H⁺ + OH^-

Answer 15

The ionic product of water Kw = [H⁺][OH^-] Always the same at a specific temperature

Answer 16

Kc = [H⁺][OH^-] / [H₂O] Water only dissociates a little bit, equilibrium lies on the left. [H₂O] is considered to have a constant value (much larger than others) Kc \* [H₂O] = Kw Kw = [H⁺][OH^-]

Answer 17

[H⁺] = [OH^-] Kw = [H⁺]²

Answer 18

Kw = 10^-14 mol² dm^-6

Answer 19

[OH^-] = [B] B = a base

Answer 20

Kw = [H⁺][OH^-] [H⁺] = Kw / [OH^-] pH = - log₁₀[H⁺]

Answer 21

pKw = -log₁₀Kw

Answer 22

Probe and digital display allowing for the measurement of pH

Answer 23

Place probe into pH meter into deionised water and adjust the reading to read 7.0 Do the same with a standard solution of pH 4 and another of pH 10 Rinse the probe with deionised water between each reading

Answer 24

Compare to the concentration to determine if strong or weak

Answer 25

Calculate moles from mass and RFM of compound Concentration of acid = moles\*1000 / volume(cm³) [HA]_inital= [HA]_equilibrium [H⁺] = 10^-pH Ka = [H⁺]²/[HA]

Answer 26

[H⁺] decreases pH increases

Answer 27

Diluted by factor of 10, increases pH by 1

Answer 28

Diluted by a factor of 10, increases pH by 0.5

Answer 29

Measure base using a pipette and put in flask with indicator Rinse burette with standard solution of acid then fill with standard solution Rough titration to an idea of rough end point, do this by adding acid to base and swirling and stopping when colour change Accurate titration done within 2cm³ of end point then drop-by-drop, and work out the amount of acid needed to neutralise the base Repeat a few times to get an average titre

Answer 30

Starts low, steep equivalence point to a high plateau ![]()

Answer 31

Starts low and steep curve to just above 7 where it plateaus ![]()

Answer 32

Starts higher (nearer to 7) and curves up to high (near 14) ![]()

Answer 33

Starts high (near 7) and curves to just above 7 ## Footnote ![]()

Answer 34

Initial pH depends on strength of the acid and small amounts of base has little impact with strong acids Vertical equivalence lines at the end point where [H⁺] ≈ [OH^-] as the acid has been neutralised, tiny increase causes big change in pH pH change is less pronounced when strong acids added to strong bases compared to strong acids added to weak bases (or strong base to weak acid) Final pH depends on strength of the base - stronger the base, the higher the final pH

Answer 35

Changes colour at the end point (neutralisation) of a reaction

Answer 36

Pick one which changes colour exactly at the end point, shown on the graph by the sharp increase Changes colour over narrow pH range entirely on the vertical part of the pH curve

Answer 37

Colour at low pH = red Colour at high pH = yellow Approx. pH of colour change = 3.1-4.4

Answer 38

Colour at low pH = colourless Colour at high pH = pink Approx. pH of colour change = 8.3 - 10

Answer 39

Either methyl orange or phenolphthalein Rapid pH over range for both

Answer 40

Methyl orange pH rapidly changes over methyl orange range but not phenolphthalein range

Answer 41

Phenolphthalein pH changes rapidly over phenolphthalein range

Answer 42

Neither methyl orange nor phenolphthalein Just use a pH meter as no indicators work

Answer 43

Titration curve for a weak acid and strong base

Answer 44

Stage of titration when half of hte acid has been neutralised For using this: half of equivalence of strong base has been added to the weak acid

Answer 45

Weak acid dissociates HA ⇔ H⁺ + A^- At half-equivalence point [HA] = [A^-] Therefore Ka = [H⁺] and pKa = pH pH at half-equivalence point is the pKa of the weak acid

Answer 46

Charts which show the colour of indicators at different pH

Answer 47

A solution that minimises changes in pH when small amounts of acid or base are added

Answer 48

Only can minimise small changes Does not work with large amounts of acid/base

Answer 49

Weak acid and conjugate base pH less than 7

Answer 50

1. Mix a weak acid with salt of conjugate base, salt dissociates into ions and the weak acid only slightly dissociates 2. Mix excess of weak acid with a strong base, all base reacts with acid, as xs of weak acid still some left in the solution

Answer 51

e.g. CH₃COOH ⇔ H⁺ + CH₃COO^- Addition of H⁺ shifts equilibrium to the left, reducing H⁺value to near original Addition of OH^- reacts with H⁺, equilibrium shifts to the right meaning H⁺ conc returns to near original

Answer 52

Weak base and one of its salts E.g. Ammonia solution and ammonium chloride

Answer 53

A basic solution that's soluble in water

Answer 54

Salt of weak base is fully dissociated in the solution NH₄Cl → NH₄⁺ + Cl^- Equilibrium set up between the ammonium ions and ammonia NH₄⁺ ⇔ H⁺ + NH₃

Answer 55

E.g. NH₄⁺ ⇔ H⁺ + NH₃ Small amount of acid added causes equilibrium to shift to the less, reducing H⁺ near to orignial value Small amount of base added reacts with H⁺, equilibrium shifts to the right due to less H⁺ so returns to original value. Can occur due to plenty of NH₄⁺ due to the salt

Answer 56

Initial quick increase in pH as base is strong Curve levels off as buffer solution of conjugate base (salt) in weak acid is formed which resists big pH change Eventually all weak acid used up and equivalence point is reached

Answer 57

Carbonic acid-hydrogencarbonate buffer system

Answer 58

H₂CO₃ ⇔ H⁺ + HCO₃^- Increase in H⁺ causes equilibrium to shift to the left, reducing it to the regular value Decrease in H⁺ causes equilibrium to shift to the right, increasing [H⁺] to regular value

Answer 59

H₂CO₃ ⇔ H₂O + CO₂ Controlled by respiration Breathing out CO₂ reduces H₂CO₃ Levels of HCO₃^- controlled by the kidneys, excess excreted in the urine

Answer 60

Salt of conjugate base is fully dissociated, [A^-]_{initial of salt} ≈ [A^-]_equilibrium HA is only slightly dissociated so assume [HA]_initial ≈ [HA]_equilibrium

Answer 61

Conjugate base doesn't only come from dissociation of the weak acid

Answer 62

Ka = [H⁺][A^-] / [HA] [H⁺] = Ka \* [HA]/[A^-] Use data value of Ka and assumptions that [HA] and [A^-] are equal to inital concentrations

Answer 63

pH = pKa + log₁₀([A^-]/[HA]) Relies on [A-]_initialof salt ≈ [A-]_equilibrium, and [HA]_initial≈ [HA]_equilibrium Allows for formation of a buffer with a specific pH

Answer 64

Calculate [H⁺] in original solution Calculate [H⁺] in diluted solution = [H⁺] \* (old/ new volume) Calculate pH using new [H⁺]

Answer 65

Temp increases, equilibrium moves to the right [H⁺] and [OH^-] increase Kw increases and pH decreases Water still neutral as [H⁺] = [OH^-] but lower pH

Answer 66

[OH^-] in dilute = [OH^-] in original \* (old/new volume) [H⁺] = Kw/[OH^-] Kw = 10^-14 at room temp

Answer 67

Calculate moles of H⁺ Calculate moles of OH^- Work out which in excess and by how many moles Calculate conc. of excess Calculate pH

Answer 68

Ka = [H⁺]²/[HA] Calculate [H⁺] and pH Remember [HA]_initial = [HA]_equilibrium

Answer 69

Calculate moles of HA Calculate moles of OH^- Calculate which is in excess If XS HA, calculate moles and conc of HA and A^-, use Ka to find [H⁺] If XS OH^-, use Kw to find [H⁺] If equal moles, pH = pKa of weak acid

Answer 70

H⁺ + OH^- → H₂O 2H⁺ + CO₃²^- → H₂O + CO₂ H⁺ + HCO₃^- → H₂O + CO₂ H⁺ + NH₃ → NH₄⁺

Answer 71

HA + OH^- → A^- + H₂O Calculate amount of A^- formed and HA, then conc Use Ka = [H⁺][A^-]/[HA]

12: Acids and Bases Flashcards

(96 cards)