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Currently I am attempting two questions on concentration but I am not sure if I am correct. Can someone help me with these questions? These are the questions:

1) Calculate the volume of pure water added to dilute $\pu{0.2 M}$ $\ce{HCl}$ solution in $\pu{250 cm3}$ to $\pu{0.12 M}$ $\ce{HCl}$ solution.

$\pu{0.2 M}$ solution contains $\pu{0.05 mol}$. $\pu{0.12 M}$ solution contains $\pu{0.03 mol}$. The difference between the two solutions is $\pu{0.02 mol}$ (Which is how much water was added), so

$$V(\ce{H2O}) = 0.02 \times (2 \times 1 + 1 \times 16) = \pu{0.36 L}$$


2) Calculate the mass of $\ce{Na}$ that will form $\pu{0.15 M}$ solution of $\ce{NaOH}$ by reacting with $\pu{400 cm3}$ of water.

I know that this method is totally wrong but I'll show you:

$$\pu{0.15 M} = \frac{n}{0.4} \implies n = 0.06$$ $$m(\ce{Na}) = 0.06 \times 23 = \pu{1.38 g}$$

where $n$ just refers to the concentration ($C = n/V$).

Edit:

Thanks to @George Tian I have made another attempt for the second question.

I have calculated the number of moles of NaOH for a 0.15M solution:

c = n/v
n = 0.4 x 0.15
  = 0.06

And I calculated how much Na is needed for this reaction:

The balanced equation is: 2 Na + 2 H20 -> 2 NaOH + H2
So, 0.06 mol of Na is needed (Due to the molar ratio)

Then I converted moles to grams

Mass = 0.06 x 23
     = 1.38 g
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closed as off-topic by Mithoron, Jan, Todd Minehardt, Jon Custer, hBy2Py Aug 25 '17 at 15:30

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To clarify, I've reworded your questions. Please comment if I changed the meaning:

  1. Calculate the volume of pure water that needs to be added to dilute a 0.2M HCl solution with a volume of 250cm3 to a 0.12M HCl solution

The first step would be to work out how many moles of HCl is in the original solution. As you have stated

c = n / v

c = 0.2M

v = 250cm3 or 0.250L.

Therefore n = 0.05m

Then work out how much water is needed to make a 0.12M solution.

c = 0.12M

n = 0.05m

Therefore v = 0.42L

To work out how much water needs to be added, subtract the original 0.25L from 0.42L, which equals 0.17L

  1. Calculate the mass of Na needed to react with 400cm3 of water to form a 0.15M NaOH solution.

As you haven't progressed much on this question, I'll give a few hints.

Write a balanced equation

Work out how many moles of NaOH is needed to make 0.15M solution

Work out how many moles of Na is needed to form that many moles of NaOH

Convert from moles to mass.

Hope this helps.

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  • $\begingroup$ Thank you. Could you please check my edited question for the second question? $\endgroup$ – kimchiboy03 Aug 25 '17 at 22:42
  • $\begingroup$ @kimchiboy03 Well done, that is correct. However, I'd use 4 significant figures for the molar mass of Na (22.99), and ensure the amount of significant figures in the final answer is correct. $\endgroup$ – George Tian Aug 25 '17 at 23:42
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There is no way to take a 0.2M solution of anything and get a 0.12M solution. Although it is trivial to take a 0.20M or even a 0.200M solution and dilute it to 0.12M. Have you ever heard of significant digits?
You state that a 0.12M solution contains 0.03 moles...so if I get a million tons of 0.12M solution or a millionth of a microgram of 0.12M solution they both contain 0.03 moles of solute?? I. don't think. so. In a dilution the number of moles of the solute do not change (of course, I'm talking about a one pot dilution, in serial dilutions, you only use part of the solutions, so the amount of solute does change). If you start with .05 moles of solute, then you will end with 0.05 moles of solute (ignoring losses).
So, you calculate how much HCl you have (moles): 0.20 x 0.250 = 0.050 moles
Then you want to convert that to a concentration of 0.12 moles/L: 0.050/x = 0.12 or x = 0.05/0.12 = 416.67 mL (during calculations, you should bring thru more digits that you need in the result). You started with 250 which means you need 416.67-250.00 = 166.67 ml or 167 ml after adjusting for significance.
When solving these kinds of problems, I generally try to convert to things which are conserved. Atoms are conserved, density and concentration and volume are not, so converting from concentration to moles means you're less likely to make an obvious error. At least, that's my experience.

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