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.

Md Ariful Islam

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Md Ariful Islam

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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 .

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

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

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