# What is the %"w/v" of "NaOH" if "4.0 g" of it was in "500 mL" and "10 mL" of it was used to get "100 mL"?

Jun 6, 2017

$\text{0.08% w/v NaOH}$, assuming the $\text{500 mL}$ is the final total volume of the starting solution.

It's a little unclear exactly what you mean, but I assume you are saying:

• $\text{4.0 g}$ of sodium hydroxide was dissolved in water and diluted to $\text{500 mL}$.
• $\text{10 mL}$ of that solution was diluted to $\text{100 mL}$.
• We want the resultant concentration in % "w/v".

We define %"w/v" as:

bb(%"w/v" = "g solute"/"mL soln" xx 100%)

We started with:

$\text{4.0 g NaOH"/"500 mL soln"xx100% = 0.8% "w/v NaOH}$,

because the $\text{mL soln}$ is for the final volume (including volume changes due to its displacement by the solute).

If we then take $\text{10 mL}$ of this solution and dilute it to $\text{100 mL}$, we have diluted it $10$-fold.

Since the volume multiplied by $10$, the concentration divided by $10$, as the same number of particles present in a larger volume has a smaller concentration.

We can more easily see this in the dilution equation, where $M$ is concentration and $V$ is volume:

${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

(0.8% "w/v NaOH")("10 mL") = M_2("100 mL")

=> M_2 = ((0.8% "w/v NaOH")(10 cancel"mL"))/(100 cancel"mL")

= color(blue)(0.08% "w/v NaOH")