Chapter 17 Notes

Chapter 17 Notes - Chapter 17 I Chapter 16 Summary A HX...

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Chapter 17 I. Chapter 16 Summary A. HX +H 2 O H 3 O + +X - Ka 1. Ka=x 2 /y-x=x 2 /y B. X - +H 2 O HX+OH - Kb 1. Kb=x 2 /y-x=x 2 /b C. KaKb=Kw II. Common-Ion Effect A. HX as an acid such as HCl B. MX is a salt with a metal such as NaCl C. HX+H 2 O H 3 O + +X - 1. MX M + +X - 2. Adding MX to pure HX in water 3. This will drive the equilibrium to the left because it is like adding X - and this will decrease the concentration of H 3 O + so the pH will increase D. Solution of .3M HC 2 H 3 O 2 and .3M NaC 2 H 3 O 2 what is the pH? 1. The NaC 2 H 3 O 2 will dissociate to form Na and C 2 H 3 O 2 2. HC 2 H 3 O 2 +H 2 O H 3 O + +C 2 H 3 O - Ka=1.8E-5 3. .3-x x .3+x 4. 1.8E-5=x(.3+x)/(.3-x)=x(.3)/.3=x 5. [H 3 O]=1.8E-5 6. pH=-log1.8E-5=4.74 7. pH of .3M solution of HC 2 H 3 O 2 =2.64 III. Buffered Solution A. Weak acids + salt of conjugate bases or vice versa B. Solutions made up like this resist changes in pH C. Blood pH varies between 7.35 to 7.45 1. pH<6.8, die 2. pH>7.8, die 3. Acidosis pH<7.35 4. pH>7.74 alkalosis D. HCO 3 - +H 3 O H 2 CO 3 +H 2 O 1. H 2 CO 3 H 2 O+CO 2 E. HX+H 2 O H 3 O+X 1. Ka=[H 3 O][X]/[HX] 2. Ka=[H 3 O]([X] o +x )/([HX] o -x )= [H 3 O][X] o /[HX] o 3. –logKa=pKa=-log[H 3 O]-log[X] o +log[HX] o 4. pKa=pH-log[X] o /[HX] o 5. pH=pKa+log([X] o /[HX] o ) F. Mixture of acid HA + salt MA
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