Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry251Example 6.12Calculate the pH of a buffer that is 0.020 M in NH 3 and 0.030 M inNH 4 Cl. What is the pH after adding 1.0 mL of 0.10 M NaOH to 0.10L of this buffer?So l u t i o nThe acid dissociation constant for NH 4 + is 5.70 × 10 –10 , or a pK a of 9.24.Substituting the initial concentrations of NH 3 and NH 4 Cl into equation6.60 and solving, we find that the buffer’s pH is0 020pH = 924 . + log . = 906 .0.030Adding NaOH converts a portion of the NH 4 + to NH 3 as a result of thefollowing reaction+ −NH + OH HO+NH4 23With a pH of 9.06, the concentration ofH 3 O + is 8.71×10 –10 and the concentrationof OH – is 1.15×10 –5 . Because bothof these concentrations are much smallerthan either C NH3 or C NH4 Cl , the approximationsleading to equation 6.60 arereasonable.Because this reaction’s equilibrium constant is so large (it is 5.7 × 10 4 ), wemay treat the reaction as if it goes to completion. The new concentrationsof NH 4 + and NH 3 areC+=NH4molNH+ −4V− mol OHtotal−3( 0. 030 M)(0.10 L) − ( 010 . M)(1.0×10 L)== 0. 029 M−3 0.10 L + 1.0×10 LCNH3molNH + mol OH3=Vtotal−−3( 0. 020 M)(0.10 L) + ( 010 . M)(1.0×10 L)== 0.021 M−3 0.10 L + 1.0×10 LSubstituting these concentrations into the equation 6.60 gives a pH ofPractice Exercise 6.120 021pH = 924 . + log . = 910 .0.029Calculate the pH of a buffer that is 0.10 M in KH 2 PO 4 and 0.050 M inNa 2 HPO 4 . What is the pH after adding 5.0 mL of 0.20 M HCl to 0.10L of this buffer. Use Appendix 11 to find the appropriate K a value.Click here to review your answer to this exerciseNote that adding NaOH increases the pHfrom 9.06 to 9.10. As we expect, adding abase makes the pH more basic. Checkingto see that the pH changes in the rightdirection is one way to catch a calculationerror.