Question #216ab

1 Answer
Apr 14, 2017

Answer:

#"0.083 L"# of concentrated #12.0M# #HNO_3#

Explanation:

This is a dilution chemistry problem. For problems like this one, use the following equation

#color(white)(aaaaaaaaaaaaaaaa)##M_1V_1 = M_2V_2#

Where
#M_1 = "molarity of solution 1"#
#V_1 = "volume of solution 1"#
#M_2 = "molarity of solution 2"#
#V_2 = "volume of solution 2"#

Whenever you dilute a concentrated stock solution, or for that matter, any solution, you are essentially changing the volume of that solution by adding water. If we look at what #"Molarity"# exactly is,

#"Molarity" = "moles"/(1" Liter solution")#

you can see when you increase the #"volume"#, you decrease the #"molarity"# (concentration). In dilution cases, you are adding water so moles of the solute do not change, only the volume of the solution changes.

#color(white)(aaaaaaaaaaaaaaaa)#

Using our equation from above, we can find the number of moles of nitric acid, #HNO_3#, from what we are given.

We will let #M_1# and #V_1# be #8.0 M# and #125" mL"# respectively. Make sure to convert #"mL"# to #"L"# first. For #M_2#, use the #12.0M#. Isolate #V_2# and solve.

  • #M_1V_1 = M_2V_2#

  • #(0.125L)(8.0M) = (12.0M)(V_2)#

  • #((0.125L)(8.0cancelM))/(12.0cancelM) = V_2#

  • #"0.083 L"= V_2#

Answer: 0.083 L