nacho's question thread

[nacho]

I figured it would be best to make a thread for my questions, so i can track them down later and see if i can still solve them.
For now, the following:

Q1. An industrail plant has a tank containing 500L of waste queuous copper(II) sulfate solution with a concentration of 0.0014 M.
i. What is the oncentration of the copper(II) ions in this solution, expressed in mg L^-1
- The concentration of copper(II) ions would be the same as the concentration of copper(II) sulfate, right? So all the question is asking me to do is convert from 0.0014 M to mg L^-1 ?
although, i have my doubts because of the follow up question..and my interpretation of the Q seems to easy...

ii. In order to discharge this solution into the sewers the plant will have to dilute this solution with sufficient water so that copper(II) ion concentration is not greater than 10 mg L^-1. What is the minimum volume of waer that the plant must add to the tank to dilute it prior to discharge?

Q2. A student dissolved 0.080g of solid sodium hydroxide in deonised water to make up 200.0 mL of solution.
i. What is the concentration of this solution?
ii. What is the pH of this solution?

Also,
I have a solubility curve for ammonium chloride (NH4Cl)
and the question asks:
i. What mass of ammonium chlroide would need to be dissolved in 500mL of deionised water to prepare a solution that was just saturated at 48 degrees celsius.

ii. what is the concentration of the ammonium ion in this solution?

Do i just find the value at which it dissolves in g/100mL at 48 degrees and then multiply this by 5?
___________________
This is all for now.
Thanks in advance

[Mao]

Q1 i, yes, that is it. Converting 0.0014M to mg/L. But be careful with the wording of the question, they're asking for mg/L of copper (II) ions only.

Q2 i, m(NaOH) --> n(NaOH) --> c(NaOH)
Q2 ii, [H+][OH-]=10^-14, pH = -log([H+])

Q3 i, your interpretation is again spot on.

[nacho]

Q1 i, yes, that is it. Converting 0.0014M to mg/L. But be careful with the wording of the question, they're asking for mg/L of copper (II) ions only.

Just to clarify, this would be 0.0014 x 63.55 = 0.089 mg/L ?
Something seems wrong, because of the follow up question..

[Mao]

Q1 i, yes, that is it. Converting 0.0014M to mg/L. But be careful with the wording of the question, they're asking for mg/L of copper (II) ions only.

Just to clarify, this would be 0.0014 x 63.55 = 0.089 mg/L ?
Something seems wrong, because of the follow up question..

0.0014 mol/L * 63.55 g/mol = 0.089 g/L = 89 mg/L

[nacho]

failed... :|
thanks

[luffy]

Q1 i, yes, that is it. Converting 0.0014M to mg/L. But be careful with the wording of the question, they're asking for mg/L of copper (II) ions only.

Just to clarify, this would be 0.0014 x 63.55 = 0.089 mg/L ?
Something seems wrong, because of the follow up question..

0.0014 mol/L * 63.55 g/mol = 0.089 g/mol = 89 mg/mol

I feel a little threatened by correcting Mao, so I am probably incorrect. But, shouldn't it be 89mg/L

[Mao]

Q1 i, yes, that is it. Converting 0.0014M to mg/L. But be careful with the wording of the question, they're asking for mg/L of copper (II) ions only.

Just to clarify, this would be 0.0014 x 63.55 = 0.089 mg/L ?
Something seems wrong, because of the follow up question..

0.0014 mol/L * 63.55 g/mol = 0.089 g/mol = 89 mg/mol

I feel a little threatened by correcting Mao, so I am probably incorrect. But, shouldn't it be 89mg/L

You are quite right! In my excuse 3PM is a bit too early for me, I would only have been awake for half an hour

[luffy]

Q1 i, yes, that is it. Converting 0.0014M to mg/L. But be careful with the wording of the question, they're asking for mg/L of copper (II) ions only.

Just to clarify, this would be 0.0014 x 63.55 = 0.089 mg/L ?
Something seems wrong, because of the follow up question..

0.0014 mol/L * 63.55 g/mol = 0.089 g/mol = 89 mg/mol

I feel a little threatened by correcting Mao, so I am probably incorrect. But, shouldn't it be 89mg/L

You are quite right! In my excuse 3PM is a bit too early for me, I would only have been awake for half an hour

Haha, don't worry Mao. You are still "Mao - the legendary" in my mind

[nacho]

He is 'mao' legendary than anyone, if i might add