SAMPLE BUFFER CALCULATIONS – FULL Answers 1. Calculate the pH of an unbuffered 0.010M acetic acid solution. CH3COOH < ====> CH3COO+ H+ I 0.010M ---------R E 0.010 -y y y Ka = 1.8 x 10-5 y = 4.2 x 10-4 M pH = 3.38 = y2 / 0.010 -y 2. Calculate the pH of a buffered 0.010M acetic acid solution. CH3COOH < ====> CH3COO+ H+ I 0.010M 0.010M ---R E 0.010 -y 0.010+ y y In a buffered sol'n, the conjugate base is also present Ka = 1.8 x 10-5 = y (0.010 + y) / 0.010 -y y = 1.8 x 10-5 M pH = 4.74 THE PRESENCE OF THE CONJUGATE NOT ONLY PREVENTS LARGE SWINGS, IT GUARANTEES A MORE NEUTRAL INITIAL EQUILIBRIUM! 3. Calculate the pH of an unbuffered 0.010M acetic acid solution in which 0.004 M of H+ is being added from an acidic source. CH3COOH < ====> CH3COO+ H+ I 0.010M ------0.004 M R -y -----------> +y +y E 0.010-y y 0.004 + y Ka = 1.8 x 10-5= y (0.004 + y) / (0.010-y) y= 4.5 x 10-5 [H] = ~0.004 pH = 2.40 This is a difference of 2.34 pH units from Question #2. 4. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of H+ is being added from an acidic source. CH3COOH < ====> CH3COO+ H+ I 0.010M 0.010M 0.004 M R +0.004 < ------------ -0.004 -0.004 see **** first!!! E 0.014 0.006 y~0 Ka = 1.8 x 10-5 **** K trial = (0.004)(0.0010) / (0.010) = 4 x 10-3 Ktrial > K Thus in the R line the equil. shifts left Ka = 1.8 x 10-5= y (0.006) / (0.0014) y= 4.2 x 10-5 (note that we assumed that [H] e was ~0. This assumption is valid compared to the intial concentrations). pH = 4.38 This is a difference of only 0.36 pH units from Question #2. 5. Calculate the pH of an unbuffered 0.010M acetic acid solution in which 0.004 M of OH- is being added from an basic source. I R E CH3COOH 0.010M -0.004 0.006 + OH- < ====> CH3COO+ 0.004 M -------0.004 -----------> +0.004 y~0 0.004 H2O X X X K =1/Kb= Ka/Kb (It is the anion hydrolysis in the reverse direction!) K = 1.8 x 109=0.004/ 0.006 (y) y= 3.7 x 10-10 (note that we assumed that [OH] e was ~0. This assumption is valid compared to the intial concentrations). [OH] = ~3.7 x 10-10 pOH = 9.43 pH = 4.57 6. Calculate the pH of a buffered 0.010M acetic acid solution in which 0.004 M of OHis being added from an basic source. CH3COOH + OH- < ====> CH3COO+ H2O I 0.010M 0.004 M 0.010M X R -0.004 -0.004 -----------> +0.004 X ****** E 0.006 y~0 0.014 X ****** Ktrial = (0.010) / (0.010)(0.004) = 2.5 X 102 Ktrial < K shifts right K =1/Kb= Ka/Kb (It is the anion hydrolysis in the reverse direction!) K = 1.8 x 109=0.014/ 0.006 (y) y= 1.3 x 10-9 (note that we assumed that [OH] e was ~0. This assumption is valid compared to the intial concentrations). [OH] = ~1.3 x 10-9 pOH = 8.89 pH = 5.11 EXTRA BUFFER QUESTIONS 1. 2. 3. 4. 3.32 5.12 0.57 g 9.13 5. I R E CH3COOH 0.010M -0.004 0.006 Answers + OH- < ====> CH3COO+ 0.004 M 0.010M -0.004 -----------> +0.004 y~0 0.014 H2O X X X ****** ****** Ktrial = (0.010) / (0.010)(0.004) = 2.5 X 102 Ktrial < Keq therefore the equilibrium shifts right K =1/Kb (It is the anion hydrolysis in the reverse direction!) K = 1.8 x 109=0.014/ 0.006 (y) y= 1.3 x 10-9 (note that we assumed that [OH] e was ~0. This assumption is valid compared to the intial concentrations). [OH] = ~1.3 x 10-9 pOH = 8.89 pH = 5.11