1) A solution 5.50*10‐2 M of lactic acid has pH = 2.195. Calculate the constant of dissociation of
lactic acid (CH3CHOHCOOH).
Answer: 8.37*10‐4
2) Ammonia, NH3, is a weak base with Kb = 1.79*10‐5. Calculate the concentration of a solution of NH3 so as its pH is 11.040.
Answer: 6.87*10‐2 M
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Lactic acid: CH3CHOHCOOH … AKA: HLac
HLac <==> H+ + Lac- ……. Ka = ???
[H+] can be determined from the pH, [H+] = 10^-pH, which also gives [Lac-]
[H+] = 0.638M
[Hlac] = 5.5×10^-2 – 0.638 …. this accounts for the HLac that dissociated
Ka = [H+] [Lac-] / [HLac]
Ka = 0.638² / (5.5×10^-2 – 0.638)
Ka = 8.37×10^-4
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An ammonia solution is basic because it reacts with water to make a dilute solution of hydroxide ion, plus ammonium ion.
NH3(aq) + H2O(l) <==> NH4+ + OH- …….. Kb = 1.79×10^-5
If the pH is 11.040, then pOH = 2.96, and since [OH-] = 10^-pOH, then [OH-] = 0.110M
Kb = [NH4+] [OH-] / [NH3]
1.79×10^-5 = 0.110² / (x – 0.110) ….. where x is the original concentration of NH3. The original NH3 concentration is reduced by the amount that reacts with water which is 0.110 moles per liter.
x = 0.0687
[NH3] = 0.0687M