How many mmols in NaOH are used? In an experiment, 5mL of 6 M NaOH is used along with 300 ml distilled water. The solution of the NaOH was titrated into a vinegar solution and it was found that 7.92 mL of NaOH was used.

My teacher wrote this down on the board: 6M NaOH/305 ml X 5ml =.1 M NaOH

1 Answer
Jul 5, 2017


#"0.8 mmoles NaOH"#


The idea here is that you're diluting the stock solution to a total volume of

#overbrace("5 mL")^(color(blue)("the volume of the stock solution")) + overbrace("300 mL")^(color(blue)("the volume of water")) = "305 mL"#

The thing to keep in mind about dilutions is that the ratio that exists between the volume of the diluted solution and the volume of the stock solution


and the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution


must be equal. You can thus say that you have--here #"DF"# is the dilution factor

#"DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock"#

In your case, the dilution factor is

#"DF" = (305 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = color(blue)(61)#

which means that the concentration of the diluted sodium hydroxide solution is

#c_"diluted" = c_"stock"/color(blue)("DF")#

#c_"diluted" = "6 M"/color(blue)(61) = "0.0984 M"#

Now all you have to do is to use the volume of the solution to find the number of millimoles of sodium hydroxide present in the sample

#7.92 color(red)(cancel(color(black)("mL solution"))) * (0.0984 * color(blue)(cancel(color(black)(10^3)))color(white)(.)"mmoles NaOH")/(color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.8 mmoles NaOH")))#

The answer must be rounded to one significant figure.

Notice that this is exactly what your teacher calculated by doing

#"6 M"/"305 mL" * "5 mL" ~~ "0.1 M"#

because this is actually the concentration of the stock solution divided by the dilution factor

#"6 M"/("305 mL"/"5 mL") = "6 M"/"305 mL" * "5 mL"#