Analytical Chem istry - DePauw University

280 Analytical Chemistry 2.0Adding HCl converts a portion of HPO 4 2– to H 2 PO 4 – as a result of thefollowing reaction2− + −HPO ( aq ) + H O ( aq ) H O() l + H PO ( aq )43 2 2Because this reaction’s equilibrium constant is so large (it is 1.59 × 10 7 ),we may treat the reaction as if it goes to completion. The new concentrationsof H 2 PO 4 – and HPO 4 2– areCC−=HPO 2 42− =HPO4molH PO2−4V+ molHCltotal−3( 010 . M)( 0. 10 L) + ( 020 . M)( 5. 0×10 L)=3010 . L + 5.0×10− LmolHPOV2−4total-mol HCl−3( 005 . M)( 0. 10 L) − ( 020 . M)( 5. 0×10 L)=3010 . L + 5.0×10− L4= 0.105M= 0.0381 MSubstituting these concentrations into equation 6.60 gives a pH of2−HPO40 038pH = 7. 199 + log [ ] = 7. 199 + log . 1 = 6. 759 ≈676.−[ HPO ]0.10524As we expect, adding HCl decreases the buffer’s pH by a small amount,dropping from 6.90 to 6.76.Click here to return to the chapter.Practice Exercise 6.13We begin by calculating the solution’s ionic strength. Because NaCl is a 1:1ionic salt, the ionic strength is the same as the concentration of NaCl; thusm = 0.10 M. This assumes, of course, that we can ignore the contributionsof Hg 2 2+ and Cl – from the solubility of Hg 2 Cl 2 .Next we use equation 6.63 to calculate the activity coefficients for Hg 22+and Cl – .2051 . ( 2) 010 .log γ2+ Hg2= − × + ×1+ 3. 3× 0. 40×010 .γ Hg+= 0.35122=−0.455. ( ) .log γ Cl−= − × − 2051 1 × 010=−01. 21+ 3. 3× 0. 3×0.10γ Cl−= 075 .Defining the equilibrium concentrations of Hg 2 2+ and Cl – in terms ofthe variable x