Understanding Neutralization Reactions and Henderson-Hasselbalch Equation
Explore the concept of neutralization in chemistry, illustrated with examples of strong acids and bases reacting to reach equilibrium. Dive into the Henderson-Hasselbalch equation to understand the relationship between pH, pKa, and concentrations of acids and bases in solution.
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Example How many ml of 0.05 N HCl are required to neutralize exactly 8g of NaOH? At the equivalent point: The no. of moles H+added = no. of moles OH-present Lacid Nacid= no. of (moles) equivalents of H+ added wtNaOH/ MwtNaOH= no. of moles of NaOH (OH-) present Lacid Nacid= wtNaOH/ MwtNaOH Lacid 0.05 = 8 / 40 Lacid= 0.2/ 0.05 = 4 L or 4000 ml.
Henderson Hasselbalch Equation HA H++ A- [H+] [A-] [HA] Ka = [H+] [A-] =[HA] Ka Ka[HA] [A-] [H+] = Multiply by log log[H+] = log Ka + log [HA] [A-] -log Ka - log [HA] [A-] pKa + log [A-] [HA] Multiply by -1 -log[H+] = pH =
Henderson Hasselbalch Equation MOH M+ + OH- [M+] [OH-] [MOH] Kb = [M+] [OH-]=[MOH] Kb Kb[MOH] [M+] [OH-] = Multiply by log log[OH- ] = log Kb + log [MOH] [M+] - log Kb - log [MOH] [MOH] Multiply by -1 -log[OH-] = [M+] pKb+ log [M+] pOH =
