#7 please Calculate the values ofK.and pK.fr phenol e, d. Calculate the values o
ID: 567058 • Letter: #
Question
#7 please Calculate the values ofK.and pK.fr phenol e, d. Calculate the values of K, and pÅ;, for the phenoxide ion a What is the pA, of the saccharinate ion, C,H,SO, . Doss a solution of sodium saccharinate in water have a Siplt of 7, or is the solution acidie 4) The pK of saccharin, HCHSO, a sweetening agent, is 11.68. or basic? e. If the pll is not 7, calculate the pH of a 0.010 M solution of sodium saccharinate in water. At 25 0.0115 M solution of codeine in water. the value of K, for codeine, a pain-killing drug, is 1.63 x 104. Calculate the pH of a 6 Methylamine, CH,NH, is a weak base. Write the chemical equation f for the equilibrium that occurs in an aqueous solution of this solute. Write the equilibrium law corresponding to K, for CH,NH,, 7) The pk of methylamine, CH,NH, is 3.36. Calculate the pKa of its conjugate acid, CH,NH, Laboratory Conclusion.(Word Count. From 250 to 500 words) Hint: Look at the objectives of the lab and write what your group concludes about this experiment. Explain your conclusion by summarizing any data that support it. You can comment about what you learned new or you are reinforcing a concept that you saw previously. You can also comment on things that you still need to improve. Start here:
Explanation / Answer
Answer
pKa = 10.64
Explanation
CH3NH2 + H2O+ - - - - - - > CH3NH3+ + OH-
Kb = [CH3NH3+] [OH-] /[CH3NH2]
pKb = - logKb
-logKb = 3.36
Kb = 1×10-3.36=4.37×10-4
Ka= Kw/Kb
= 1.00×10-14/4.37×10-4
= 2.29×10-11
pKa = - log(Ka) = - log(2.29×10-11) = 10.64
pKa is obtained also from simple way
pKa + pKb = 14
pKa = 14 - 3.36 = 10.64
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