Question

In a typical analysis, 15 ml of an aqueous solution containing an unknown amount of acetylcholine had a ph of 7.65. When incubated with acetylcholinesterase, the ph of the solution decreased to 6.87. Assuming there was no buffer in the assay mixture, determine the number of moles of acetylcholine in the 15 ml sample.

Answer 1

pH of the acetyl choline solution before incubation = 7.65

$[H_{3}O^{+}]=10^{-7.65}=2.24*10^{-8}M$

pH of the solution after incubation = 6.87

$[H_{3}O^{+}]=10^{-6.87}=1.35*10^{-7}M$

The difference in concentration of hydronium ion before and after incubation

=$1.35*10^{-7}M$-$2.24*10^{-8}M$=$1.126*10^{-7}M$

This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.

Therefore the moles of acetylcholine = $15 mL * \frac{1L}{1000mL}*\frac{1.126*10^{-7}mol }{L}=1.689*10^{-9}mol$