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Author Question: A 25.0 mL sample of 0.150 mol L-1 butanoic acid is titrated with a 0.150 mol L-1 NaOH solution. What ... (Read 83 times)

michelleunicorn

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Question 1

What is the pH of the resulting solution if 45 mL of 0.432 mol L-1 methylamine, CH3NH2, is added to 15 mL of 0.234 mol L-1 HCl? Assume that the volumes of the solutions are additive. Ka = 2.70 × 10-11 for CH3NH3+.
◦ 2.77
◦ 11.23
◦ 4.09
◦ 9.91

Question 2

A 25.0 mL sample of 0.150 mol L-1 butanoic acid is titrated with a 0.150 mol L-1 NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of butanoic acid is 1.5 × 10-5.
◦ 4.88
◦ 4.55
◦ 4.77
◦ 1.34
◦ 3.08


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Marked as best answer by michelleunicorn on Nov 18, 2019

dellikani2015

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michelleunicorn

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Reply 2 on: Nov 18, 2019
YES! Correct, THANKS for helping me on my review


debra928

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

 

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