patino_chm2046_chapter16

Base take acid base ph pka log to remove ve sign conj

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Unformatted text preview: tion: equation: 1 [Acid] 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 The pK acid and then adjusted by the ratio of acid and acid conjugate base. conjugate pH of Buffer (Ka) Q: What is the pH of a buffer that has [HOAc] = 0.700 Q: M and [OAc-] = 0.600 M? and 0.600 HOAc + H2O ⇄ OAc- + H3O+ HOAc + -5 -5 [H3O ] = 2.1 × 10 and pH = 4.67 2.1 pH K = 1.8 × 10 a [Conj. base] pH = pKa + log [Acid] 0.600 pH = − log (1.8 × 10 ) + log 0.700 −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” approximation 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 0.095 Cl. Equilibrium: NH3 + H2O ⇄ NH4+ + OH− Equilibrium: Solution Solution [acid] pOH = pKb + log [base] 0.095 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 The If you have a buffer made of a weak acid, HA, and a salt If containing its conjugate base, A−, such as KA (K+ = containing potassium ion), I recommend you to work with the following equilibrium: following HA + H2O ⇄ H3O+ + A− Ka, K of acid HA. 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 If containing its conjugate acid, BH+, such as BHCl (Cl− = containing chloride ion. B is not boron element), I recommend...
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