Question

2. Do the following calculations (a) Using the concentration (1 M) and volume (10 ml) of the initial acid provided, calculate the initial amount (in mole) of acid in each dialysis bag. (b) Using the final pH (at the end of the experiment) inside each bag and the volume (-10 m) calculate the final amount (in mole) of acid remaining in each dialysis bag. (c) Using the final pH of the 1000 ml water, calculate the amount of acid that has dialyzed out from the dialysis bag. (d) Using the final pH of each 333 ml water in the small beaker, calculate the amount of acid that has dialyzed out from the dialysis bag each time and the sum of the three steps of dialysis. What do you conclude from this experiment? Which method of dialysis is better: one step using 1000 ml water or three steps using 333 ml three times? Explain your answer.

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Answer #1

The formula of calculation of mole is MOLARITY *Volume of soln in L

For example for part A Molarity is 1M and Volume is 10 ml or 1/100 L so moles

Of acid is 1*1/100 =10^(-2) mol

For the second part you can calculate Molarity of H plus Ion from PH of the solution

PH is equal to -log[H+], the concentration of H plus is equivalent to concentration of acid from this you can calculate moles of acid

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