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Unformatted text preview: tion:
pH = pKa - log
[Conj. base] take [acid]
[base] pH = pKa + log to remove − ve sign [Conj. base]
[Acid] The pH is determined largely by the pKa of the
acid and then adjusted by the ratio of acid and
conjugate pH of Buffer (Ka)
Q: What is the pH of a buffer that has [HOAc] = 0.700
M and [OAc-] = 0.600 M?
0.600 HOAc + H2O ⇄ OAc- + H3O+
-5 [H3O ] = 2.1 × 10 and pH = 4.67
K = 1.8 × 10
a [Conj. base]
pH = pKa + log
pH = − log (1.8 × 10 ) + log
−5 pH = 4.74 − 0.067 = 4.67 What is the pH of a buffer that is 0.50 M NH3
(pKb = 4.75) and 0.20 M NH4Cl?
find the pKb if given Kb
Assume the [B] and [HB+]
equilibrium concentrations are
the same as the initial
Substitute into the HendersonHasselbalch Equation Base
Form, find pOH
Check the “x is small”
Calculate pH from pOH NH3 + H2O ⇄ NH4+ + OH− [HB+ ] pOH = pK b + log [B] ( 0.20 ) pOH = 4.75 + log = 4.35 ( 0.50 ) [OH − ] = 10 -pOH = 10 -4.35 = 4.47 × 10 −5
4.47 × 10 −5
× 100% = 0.0089% < 5%
0.50 pH + pOH = 14.00
pH = 14.00 - 4.35 = 9.65 16 pH from using Kb
pH Q: What is the pH of a buffer that has
What 0.12 mol of NH 3 and 0.095 mol NH4Cl. Kb = 1.8 × 10-5
Equilibrium: NH3 + H2O ⇄ NH4+ + OH−
pOH = pKb + log
pOH = 4.74 + log
0.12 pOH = 4.74 +( - 0.10) pOH = 4.64
pH = 14.00 – 4.64 = 9.36 The fundamental equations of buffers
If you have a buffer made of a weak acid, HA, and a salt
containing its conjugate base, A−, such as KA (K+ =
potassium ion), I recommend you to work with the
HA + H2O ⇄ H3O+ + A− Ka, K of acid HA.
Any of the four, [HA], [H3O+], [A−], or Ka may be unknown,
whereas the others are given.
If you have a buffer made of a weak, B, and a salt
containing its conjugate acid, BH+, such as BHCl (Cl− =
chloride ion. B is not boron element), I recommend...
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