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

Answer:

#"0.8 mmoles NaOH"#

Explanation:

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

#V_"diluted"/V_"stock"#

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

#c_"stock"/c_"diluted"#

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"#