ways that WA titration differs from SA titration
  • The H+ of HCl (a strong acid) reacts with the base member of the buffer (CH3COO−). Thus, [CH3COO−] decreases and [CH3COOH] increases.
  • Yes, the pH at the equivalence point will decrease.
  • The pH at the equivalence point increases (becomes more basic) as the acid becomes weaker.
  • 1. WA has higher initial pH than SA2. pH change near eq point is smaller for WA3. the pH @ eq point is >7 for WA.
Lewis acid
  • donates electron pair
  • accepts electron pair
  • neutralization reaction
  • precipitate forms
Q = Ksp
  • more solid can dissolve, so no precipitate
  • deals with solubility?
  • HA <=> H+ + A-Ka = [H+][A-] / [HA]
  • @ eq and solution is saturated
What is the pH at the equivalence point when 0.10 M HNO3 is used to titrate a volume of solution containing 0.30 g of KOH?
  • pH = 7
  • Ksp = 4x^3
  • HF, NaF
  • CaF2
buffers act to
  • HA <=> H+ + A-Ka = [H+][A-] / [HA]
  • keep solution at a relatively constant pH
  • A solution of 100 mL of water with 20 mL of 0.1 M NaOH added.
  • The volume of added base needed to reach the equivalence point remains unchanged.
If the acetic acid being titrated here were replaced by hydrochloric acid, would the amount of base needed to reach the equivalence point change?
  • Concentrated HBr
  • will not change
  • The pH decreases.
  • greater than 7
CH3COOH <=> H+ + CH3COO-what will addition of CH3COO- shifts equilibrium to which side?
  • NH3(aq)+H2O(l)⇌OH−(aq)+NH4+(aq)
  • K sp = [Ca2+]3[PO4 3-]2
  • The pH at the equivalence point increases (becomes more basic) as the acid becomes weaker.
  • to the left to lower the [H+]
The Henderson-Hasselbalch equation, _________allows you to calculate _______
  • pH=pKa+log [base]/[acid] the pH of a buffer.
  • The OH− of NaOH (a strong base) reacts with the acid member of the buffer (CH3COOH), abstracting a proton. Thus, [CH3COOH] decreases and [CH3COO−] increases.
  • A solution of 100 mL of water with 20 mL of 0.1 M NaOH added.
  • The volume of added base needed to reach the equivalence point remains unchanged.
Titration of a SA with a SB
  • pH = 7 at eqas you add more base, pH levels off **look at graphs
  • high pH to pH = 7 at eq --> low pH
  • where amount of base = amount of acid
  • can be used to find the equivalence point in a titration
Enter the Ksp expression for AC4(s) in terms of [A] and [C].
  • pH = pKa + log ([B]/[A])
  • K sp = [Ca2+]3[PO4 3-]2
  • greater than 7
  • Ksp = [A][C]4
You can only make buffers with
  • where amount of base = amount of acid
  • equal capacity
  • a weak acid with its conjugate base or by mixing a weak base with its conjugate acid.
  • WA or WB
Enter the the Ksp expression for C2D3 in terms of the molar solubility x.
  • pH = pKa + log ([B]/[A])
  • K sp = [Ca2+]3[PO4 3-]2
  • Ksp = [A]^2[B]^3
  • Ksp = ((2x)^2)((3x)^3)
titration of WA with SB
  • Yes, the pH at the equivalence point will decrease.
  • high pH to pH = 7 at eq --> low pH
  • can be used to find the equivalence point in a titration
  • @ eq, pH > 7as more base is added, pH levels off. same as for SA.
What happens when HCl is added to this buffer?
  • Yes, the pH at the equivalence point will decrease.
  • can be used to find the equivalence point in a titration
  • The H+ of HCl (a strong acid) reacts with the base member of the buffer (CH3COO−). Thus, [CH3COO−] decreases and [CH3COOH] increases.
  • The volume of added base needed to reach the equivalence point remains unchanged.
What is the solubility-product expression for the equilibrium established when calcium phosphate, Ca3(PO4)2(s), is added to pure water at 25°C?
  • pH = pKa + log ([B]/[A])
  • K sp = [Ca2+]3[PO4 3-]2
  • Ksp = [A]^2[B]^3
  • pH = 7
what can indicators be used for?
  • HA <=> H+ + A-Ka = [H+][A-] / [HA]
  • where amount of base = amount of acid
  • a weak acid with its conjugate base or by mixing a weak base with its conjugate acid.
  • can be used to find the equivalence point in a titration
"ideal buffers" cause
  • neutralization reaction
  • Ksp = [A][C]4
  • equal capacity
  • donates electron pair
Titration of SB with SA
  • @ eq, pH > 7as more base is added, pH levels off. same as for SA.
  • where amount of base = amount of acid
  • HA <=> H+ + A-Ka = [H+][A-] / [HA]
  • high pH to pH = 7 at eq --> low pH
Adding a SA or SB =
  • donates electron pair
  • pH = pKa + log ([B]/[A])
  • equal capacity
  • neutralization reaction
Lewis base
  • pH = pKa + log ([B]/[A])
  • precipitate forms
  • donates electron pair
  • @ eq and solution is saturated
How does the pH at the equivalence point change as the acid being titrated becomes weaker?
  • The pH at the equivalence point increases (becomes more basic) as the acid becomes weaker.
  • The pH decreases.
  • The H+ of HCl (a strong acid) reacts with the base member of the buffer (CH3COO−). Thus, [CH3COO−] decreases and [CH3COOH] increases.
  • The volume of added base needed to reach the equivalence point remains unchanged.
buffers make:
  • between 6 and 8pH range is usually +- 1 unit from pKa
  • mixture of WA + salt of CB ormixture WB + salt CA+SA and neutralize WB or+SB and neutralize WA
  • HA <=> H+ + A-Ka = [H+][A-] / [HA]
  • The volume of added base needed to reach the equivalence point remains unchanged.
titrations of polypro tic acids
  • where amount of base = amount of acid
  • can be used to find the equivalence point in a titration
  • HA <=> H+ + A-Ka = [H+][A-] / [HA]
  • neutralization reaction
Enter the the Ksp expression for the solid AB2 in terms of the molar solubility x.
  • Ksp = [A]^2[B]^3
  • pH = 7
  • Ksp = 4x^3
  • Ksp = [A][C]4
buffer capacity
  • a weak acid with its conjugate base or by mixing a weak base with its conjugate acid.
  • can be used to find the equivalence point in a titration
  • Yes, the pH at the equivalence point will decrease.
  • between 6 and 8pH range is usually +- 1 unit from pKa
@ 1:1 ratio, pH =
  • donates electron pair
  • pKa because log (1/1) = 0
  • where amount of base = amount of acid
  • neutralization reaction
Buffer solutions can be produced by mixing
  • The pH at the equivalence point increases (becomes more basic) as the acid becomes weaker.
  • a weak acid with its conjugate base or by mixing a weak base with its conjugate acid.
  • A solution of 100 mL of water with 20 mL of 0.1 M NaOH added.
  • mixture of WA + salt of CB ormixture WB + salt CA+SA and neutralize WB or+SB and neutralize WA
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