Acids & Bases: Maths
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How to calculate/define pH
-log[H+]
How to calculate [H+] from pH
1 x 10^-pH
Finish the reaction and determine which is an acid and a base HCl (g) + H2O (l) ->
HCl (g) + H2O (l) -> H3O+ (aq) + Cl- (aq) HCl = Acid H2O = Base
Water partial dissociation reaction
H2O (l) ⇌ H+ (aq) + OH- (aq)
What is Kw derived from
H2O being a constant value as it hardly ionises in aq solution
Kw equation
Kw = [H+][OH-]
Value of Kw at 25C or 298K
1x10^-14
Calculate the pH of 2.20moldm-3 HNO3
[acid] = [H+] [HNO3] = [H] pH = - log(2.20) = -0.34
What does it mean that moles of acid = moles of H
[Acid] = [H+]
What to for diprotic acids when calculating pH or H+ concentratioj
pH = Times calculated [H+] by 2 [H+] = Divide calculated [acid] by 2
As [H+] and [OH-] are the same mole values, how can the Kw equation be rewritten?
Kw = [H+]^2
Effect of increasing temperature (above 298K) on Kw and pH of water
[H+] increases and Kw increases (above 1x10^-14) pH decreases below 7
Effect of decreasing temperature (below 298K) on Kw and pH of water
[H+] decreases and Kw decreases (below 1x10^-14) pH increases above 7
What is the pH of 0.65moldm-3 NaOH at standard conditions?
[NaOH] = [OH-] = 0.65 kW = [H+][OH-] [H+] = kW / [OH-] 1x10^-14 / 0.65 = 1.54x10^-14 pH = -log(1.54x10^-14) = 13.81
How could you work out pH from pOH?
pOH = -log[OH-] 14 = pH + pOH
pH changes during dilution: What is the pH change if 20cm3 0.1moldm-3 HCl has 30cm3 water added?
[H+] = 0.1 pH = -log(0.1) = 1.00 Adding water does not change HCl moles moles = 0.1 x 20/1000 = 0.002 New volume = 20 + 30 = 50cm3 = 0.05dm3 HCl new moles = 0.002 / 0.05 = 0.04 pH = -log(0.04) = 1.40
What is Ka? (Association constant)
Dissociation constant for a weak acid Larger Ka = stronger acid
What 2 equations can be formed from HA + H2O ⇌ H3O+ + A-
Ka = [H+][A-] / [HA] Ka = [H+]^2 / [HA]
Route to find the pH of weak acids
Ka = [H+]^2 / [HA] [H+]^2 = Ka x [HA] [H+] = √Ka x √[HA] pH = -log[H+]
2 equations to calculate Ka into pKa
pKa = -log(Ka) Ka = 10^-pKa
pKa and Ka value of a strong and weak acid
Strong acid = Larger Ka = Smaller pKa Weak acid = Smaller Ka = Larger pKa
Equation process of of reacting an acid and alkali of different volumes: Acid in excess
[H+] = Excess H+ moles / new volume pH = -log[H+]
Equation process of of reacting an acid and alkali of different volumes: Base in excess
[H+] = Excess OH- moles / new volume [H+] = Kw / [OH-] pH = -log[H+]
Equation process of of reacting an acid and alkali of different volumes: Weak acid in excess
[H+] = Ka x ([HA] / [A-]) [H+] = Ka x (excess HA mol / A- mol) pH = -log[H+]