Question

What is the volume of 12.0 M NaOH needed to prepare 5.00 L of a 2.50 M NaOH solution? What volume of water (solvent) is required to make this dilution? Please show work

Answer 1

Approximately
`1.04\; \rm L` of the
`12.0\; \rm M`
`\rm NaOH` solution is needed. Approximately another
`3.96\; \rm L` of water will also be required.

Assumption: the volume of these
`\rm NaOH` solutions does not depend on the quantity of
`\rm NaOH\!` (the solute) that each of them contain.

Step-by-step explanation:

Calculate the number of moles of
`\rm NaOH` formula units in that
`5.00\; \rm L` of
`2.50\; \rm M`
`\rm NaOH\!` solution:

`n(\mathrm{NaOH}) = c\cdot V = 5.00\; \rm L * 2.50\; \rm mol \cdot L^(-1) = 12.5\; \rm mol`.

Calculate the volume of a
`12.0\; \rm M`
`\rm NaOH` with that many
`\rm NaOH\!` formula units:

`\displaystyle V = (n)/(c) = (12.5\;\rm mol)/(12.0\; \rm mol \cdot L^(-1))\approx 1.04\;\rm L`.

That should be the volume of the
`12.0\; \rm M`
`\rm NaOH` solution needed to prepare that
`5.00\; \rm L` of
`2.50\; \rm M`
`\rm NaOH\!` solution. However,
`1.04\; \rm L` corresponds to only about one-fifth the volume of a
`5.00\; \rm L\!` solution. The difference in volume should be filled with pure water:

`\begin{aligned}V(\text{water}) &\approx 5.00\; \rm L - 1.04\; \rm L = 3.96\; \rm L\end{aligned}`.