Chemistry, 31.03.2020 00:57 jettskii214
9. Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.200-M methylamine, CH3NH2(aq), with 0.200 M HCl(aq): (a) before addition of any HCl(aq) (b) after addition of 17.5 mL of HCl(aq) (c) after addition of 34.9 mL of HCl(aq) (d) after addition of 35.0 mL of HCl(aq) (e) after addition of 35.1 mL of HCl(aq)
Answers: 3
![K_{b} =\frac{[CH_{3}NH_{3}]+ [OH-] }{[CH_{3}NH_{2} ] }\\4.6x10^{-4} =\frac{x*x}{0.2-x} \\x=0.00936 M](/tpl/images/0572/2081/59e94.png)
![[acid]=\frac{M_{2}V_{2} }{V_{1}+V_{2} } =\frac{0.2*17.5}{35+17.5} =0.0667M\\[base]=\frac{M_{1}V_{1} }{V_{1}+V_{2} } =\frac{0.2*35}{35+17.5} =0.13M](/tpl/images/0572/2081/75f14.png)
![pOH=pK_{b} +log(\frac{[salt]}{[base]} )=3.34+log(0.0667/0.066)=3.34\\pH=14-3.34=10.66](/tpl/images/0572/2081/0a576.png)
![[acid]=\frac{0.2*34.9}{35+34.9} =0.0998M\\\\ [base]=\frac{0.2*35}{35+34.9} =0.1M\\base-left=0.1-0.0998=0.0002M\\pOH=3.34+log(0.0998/0.0002)=6.04\\pH=14-6.04=7.96](/tpl/images/0572/2081/d31c6.png)
![[acid]=\frac{0.2*35}{35+35} =0.1M\\[base]=\frac{0.2*35}{35+35} =0.1M\\\\pH=7-\frac{1}{2} (pK_{b}+ log(c))=7-\frac{1}{2}(3.34+log(0.1))=5.83 \\](/tpl/images/0572/2081/b3303.png)
![[acid]=\frac{0.2*35.1}{35+35.1} =0.1M\\[base]=\frac{0.2*35}{35+35.1} =0.0998M\\acid-left=0.1-0.0998=0.0002M\\pH=-log(0.0002)=3.7\\](/tpl/images/0572/2081/1cd67.png)
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9. Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.200-M methylami...
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