Question
You have 6.83 g of Alanine-HCl (MW= 125 g/mol). It is dissolved in 50 mL of water. You take 40 mL of the solution and add 15.0 mL of 6M NaOH. What is the pH?

H Alanine • Cl(s) H Alanine (aq) + H2O(1) HAlanine(aq) + H2O(1) = HAlanine (aq) + cr (aq) HAlanine(aq) + H30*(1) Alanine (
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Answer #1

CON+ cu-coornice clit Nu-cu-coon cu H2O (1) El alu.-cu-COOH - 4 + Mucu-coon CM3 Ka: 4.53X103 (M₂ NUz-Cu-coon thio - Nu-Cu-coonitial pH = -log [hi] from Eg nitial pn = -log[7.03589653X20-27 10 15263056 -7 Now, Conc f Naon= 6 m . Ud of Daou added - 13ans Ira 55X10-3 + NH₃-CH2-CO + He ce-nut Cup Coon one of on due to hydrolysis of NH₃-cu-c00a (Krocla 2 (= 0.043712S ¥10-1400pH=12.736

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