Textbook 6B.11a

[Nancy Li 1C]

I understand how to calculate part i but could someone explain part ii? Thanks!

[Meghan Terrell 1B]

I had this same question and someone posted a really helpful response

viewtopic.php?f=49&t=89234

[Mia Yamada 1J]

Hi Nancy!

Part (ii) asks us for the molar concentration of hydroxide ions in the original solution. We know from the problem that 5.00 mL of the solution was transferred to another volumetric flask and diluted to 500.0 mL. And we now know that the molar concentration of hydroxide ions in the diluted solution is 1.78x10^-1M, which we found in part (i). So, what we need to do is work backward to get to our original solution.

Let's start with trying to find the number of moles in the diluted solution since we already know the volume of diluted solution (500.0mL) and molarity of the diluted solution (1.78x10^-1M).

Using M (molarity) = mol/L, the diluted solution would be 1.78x10^-1M=(Xmol)/(.5L). So, the number of moles in the diluted solution is 0.089mol NaOH.

Then, since it asks for the molar concentration (aka molarity) of the original NaOH solution, we would take 0.089mol (which would technically be the number of m0oles of NaOH before dilution) and divide it by 0.005L (the volume of solution transferred to another volumetric flask before dilution).

The original molar concentration is 17.8M NaOH, and since NaOH is a strong base, it completely dissociates in water, so [NaOH]=[OH-]. The final answer is 17.8M OH-.

Hope this helps!

[Crystal Ma 2J]

for part ii you have to use M1V1=M2V2! doing so, you will get M1=18 mol/L