Question

You have a 15 mL sample of acetylcholine (a neurotransmitter) with an unknown concentration and a pH of 7.72. You incubate this sample with the enzyme acetylcholinesterase to convert all of the acetylcholine to choline and acetic acid. The acetic acid dissociates to yield acetate and hydrogen ions. At the end of the incubation period, you measure the pH again and find that it has decreased to 6.31. Assuming there was no buffer in the assay mixture, determine the number of nanomoles of acetylcholine in the original 15 mL sample.
Remember to use significant figures in your answer.
acetycholine in original sample:

Answer 1

The number of nanomoles of acetylcholine in the original 15 mL sample will be "7.02".

Given:

→ $P^H = -log_{10} [H^+]$

According to the question, we get

→ $P^H = 7.72 = -log_{10}[H^+]$

→ $[H^+] = 10^{-7.72}$

           $= 1.90\times 10^{-8} \ M$

The pH decreased will be:

→ $P^H = 6.31 = -log_{10} [H^+]$

→ $[H^+] = 10^{-6.31}$

           $= 4.87\times 10^{-7}$

now,

The difference in $[H^+]$ will be:

= $(4.87-0.190)\times 10^{-7}$

= $4.68\times 10^{-7} \ M$

then,

→ The number of moles will be:

= $Molarity \ of \ the \ solution\times Volume \ of \ the \ solution$

= $4.68\times 10^{-7}\times 0.015$

= $7.02\times 10^{-9} \ moles$

As we know that 1 mole = 10⁹ nanomoles.

hence,

= $7.02\times 10^{-9}\times 10^{9}$

= $7.02 \ nanomoles$ (acetylcholine)

Thus the above solution is correct.

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