What is the decay rate of Mg−28 in the solution after 4.00 days?

A 238 mL sample of an aqueous solution contains 3.05 %MgCl2 by mass. Exactly one-half of the magnesium ions are Mg−28, a beta emitter with a half-life of 21 hours. Given that half the ions are Mg−28, the molar mass of this sample is 96.9 g/mol. Assume a density of 1.02 g/mL for the solution.

1 Answer
Truong-Son N.
May 20, 2018

I got a decay rate after #"4 days"# to be #2.22 xx 10^(-4) "M/hr"#. Did you get an initial rate of #5.30 xx 10^(-3) "M/hr"#?

DISCLAIMER: LONG ANSWER!

From the density of the solution, we can find its mass:

#"1.02 g soln"/cancel"mL soln" xx 238 cancel"mL soln" = "242.76 g soln"#,

And the solution contains #3.05%"w/w"# #"MgCl"_2#. Therefore, it contains

#242.76 cancel"g soln" xx "3.05 g MgCl"_2/(100 cancel"g soln") = "7.404 g MgCl"_2#.

Using the molar mass of the sample as #"96.9 g/mol"#, and the assumption that half of the #"Mg"^(2+)# atoms are #"Mg-28"#:

#7.404 cancel("g MgCl"_2) xx cancel("1 mol MgCl"_2)/(96.9 cancel("g MgCl"_2)) xx cancel("1 mol Mg"^(2+))/cancel("1 mol MgCl"_2) xx ("1 mol "_(12)^(28) "Mg"^(2+))/(2 cancel("mol Mg"^(2+))#

#= "0.0382 mols "_(12)^(28) "Mg"^(2+)#

Knowing that its half-life is #"21 hr"#, and that #4# days have passed (or #"96 hr"#), first we find the amount leftover.

Using the integrated rate law for first-order decay in terms of mols:

#ln(n_(""_(12)^(28) "Mg"^(2+))) = -kt + ln(n_(""_(12)^(28) "Mg"^(2+),0))#

The first-order half-life is given by

#t_"1/2" = (ln2)/k#,

thus giving the rate constant as

#=> k = (ln2)/t_"1/2"#.

So:

#ln(n_(""_(12)^(28) "Mg"^(2+))) = -(ln2)(t/t_"1/2") + ln(n_(""_(12)^(28) "Mg"^(2+),0))#

#= -(ln2)("96 hr"/"21 hr") + ln("0.0382 mols "_(12)^(28) "Mg"^(2+))#

#= -6.43#

As a result, this many mols of #""_(12)^(28) "Mg"^(2+)# was left after #4# days:

#n_(""_(12)^(28) "Mg"^(2+)) = e^(-6.43) = "0.00161 mols"#

This was in #"238 mL"#, so its concentration at #t = "96 hr"# is:

#[""_(12)^(28) "Mg"^(2+)] = "0.00161 mols"/"0.238 L" = "0.00672 M"#

Finally, we can write the first-order rate law to find the decay rate at time #t#:

#r(t) = k[""_(12)^(28) "Mg"^(2+)] = (ln2)/(t_"1/2")[""_(12)^(28) "Mg"^(2+)]#

At #t = "96 hr"#, we use the leftover concentration get:

#color(blue)(r(96)) = (ln2)/("21 hr")("0.00672 M")#

#= color(blue)(2.22 xx 10^(-4) "M/hr")#

What was the initial rate of decay? Did you get #5.30 xx 10^(-3) "M/hr"#?

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