CHEM 108 1st Edition Lecture 19Chapter 16 Acid–Base and Solubility Equilibrium continuedThe pH Scale- Quantitative measure of acidity: pH log [H+ ], where [H+ ] is in M o Strong acids have a lower pH since strong acids fully dissociate in water so it increases the [H+] concentration, and decreases the pHo lower pH, stronger acid (weaker base) o higher pH, weaker acid (stronger base) o Usually, pH defined for aqueous solutions- pOH = −log[OH-]- Kw = 1 × 10−14 = [H+ ][OH− ]o Log form of Kw : −log(Kw )= −log([H+] [OH− ]) - Concentration of [H+] = [OH− ] Solution is neutral- 14.00 = pH + pOHpK: A way of expressing the strength of an acid or base - pKa = −log(Ka ), Ka = 10−pKao The stronger the acid, the smaller the pKa o larger Ka = smaller pKa » because it is the –log- pKb = −log(Kb ), Kb = 10−pKb o The stronger the base, the smaller the pKb o larger Kb = smaller pKbStrong Acids and Bases- Add strong acid to water: HA + H2O A- + H3O+ o Equilibrium lies far to the right [H3O+ ] [HA] pH = log[H+ ] = log[HA]- Calculate the pH for a 0.0500 M HCl solution.[H+]= 0.05M pH=-log[0.05M] = 1.3o Since it is a strong acid it disassociates completely and the [H+] is equal to the concentration of the acid- Add strong base to water: B + H2O BH+ + OH o Equilibrium lies far to the right o [OH ] = [B] o pOH = log[OH] = log[B] pH = 14 pOH (at 25 C)- Example: what is the pH of a 6 M solution of NaOH at 25 ºC? pOH= -log[6M]= -0.778pH = 14 (-0.778) = 14.778 at 25 ºCThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- What is the pH of a solution made by mixing 10.0 ml of 1.0 M HCl with 10.0 ml of 2.0 M NaOH at 25 ºC?HCl (aq) + NaOH (aq) H2O (aq) + NaCl (aq)moles of HCl = 0.01L* 1.0M= 0.01mmoles of NaOH= 0.01L* 2.0M= 0.02m0.02m- 0.01m= 0.01 moles of NaOH o (the solution neutralizes base with acid and what’s left is 0.01 moles of NaOH, so the solution is basic)New concentration of NaOH = (0.01moles)/(0.02 liters) = 0.5 MpOH= -log[0.5M]= 0.3pH = 14 (0.3)= 13.7 which also shows that the solution is really
View Full Document
Unlocking...