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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 23.1 5.111 Lecture Summary #23 Acid/Base Equilibrium Continued Topic: Titrations From Friday’s handout Acid buffer action : The weak acid, HA, transfers protons to OH- ions supplied by strong base. The conjugate base, A- , of the weak acid accepts protons from the H 3 O + ions supplied by a strong acid. A strong acid and the salt of its conjugate base don't make a good buffer. Why? Base Buffer Example: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH- (aq) When strong acid is added, NH 3 accepts protons from incoming acid to make NH 4 + . When strong base is added, NH 4 + donates a proton to form NH 3 and H 2 O. pH remains the same. Base buffer action : The weak base, B, accepts protons supplied by strong acid. The conjugate acid, BH + , of the weak base transfers protons to the OH- ions supplied by a strong base. A buffer is a mixture of weak conjugate acids and bases that stabilize the pH of a solution by providing a source or sink for protons. Sample Buffer Problem: Suppose 1.00 mol of HCOOH and 0.500 mol of NaHCOO are added to water and diluted to 1.0 L. Calculate the pH. (K a = 1.77 x 10-4 ) HCOOH + H 2 O H 3 O + + HCOO- initial molarity 1.00 0 0.500 change in molarity -x +x +x equilibrium molarity 1.00 -x +x 0.500 + x K a = 1.77 x 10-4 = Using approximation that x is small compared to 1.00 and 0.500, x= Check assumption pH = Now - what if 0.100 mol of a strong acid (HCl) had been included in the 1.0 L solution. Because 0.100 mol of HCl reacts with equal number of moles of HCOO- to form equal moles of HCOOH: For HCOO- , 0.500 mol - 0.100 mol = 0.400 mol [HCOO- ] = 0.400 mol/1.0 L =0.400 M For HCOOH, 1.00 mol + 0.100 mol = 1.10 mol [HCOOH] = 1.10 mol/1.0 L =1.10 M 23.2 HCOOH + H 2 O H 3 O + + HCOO- initial molarity change in molarity ______ equilibrium molarity K a = 1.77 x 10-4 = Using approximation that x is small compared to 1.10 and 0.40, x= Check assumption (5% rule) pH = 3.31 So addition of 0.10 mol of strong acid only changed pH from 3.45 to 3.31 Designing a Buffer One must consider the relationship between the ratio of [HA] to [A- ], the pK a , and the desired pH. [H 3 O + ][A - ] HA (aq) + H 2 O H 3 O + (aq) + A- (aq) K a = [HA] Rearrange: [H 3 O + ] = K a x [HA] [A- ] [HA] Take logarithms of both sides: log [H 3 O + ] = log K a + log [A- ] [HA] Multiply by (-) : -log [H 3 O + ] = -log K a - log [A- ] That is: pH = pK a - log eq [A- ] [HA] The values of [HA] and [A- ] in the equation are at equilibrium. However, a weak acid HA typically 23.2 HCOOH + H 2 O H 3 O + + HCOO- initial molarity change in molarity ______ equilibrium molarity K a = 1.77 x 10-4 = Using approximation that x is small compared to 1.10 and 0.40, x= Check assumption (5% rule) pH = 3.31 So addition of 0.10 mol of strong acid only changed pH from 3.45 to 3.31 Designing a Buffer One must consider the relationship between the ratio of [HA] to [A...
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