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Question 1 5 pts If 27.2 mL of 0.102 M acid with a pKg of 5.44 is titrated with 0.1 M NaOH solution, what is the pH of the ti
Question 3 5 pts If 23.5 mL of O.103 M acid with a pK, of 4.15 is titrated with 0.105 M NaOH solution, what is the pH of the
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Answer #1

Answer to the problem Question 1 Ha = 27. 2 ml of 0.102m; Moles of HA = 2.78x16 NaOH = 11.3 ml of 0.4M ; Moles of NaoH = 11.3-5 xxx 4, 68X10 (0.102-1) : 0.102-1 n 0.102 = 4.68x105 0.102 - x² = 4.788106 x = 2.194103 x = [H3 04] = 2.19 x1ő % M pH = -lo

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