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### Topic: 5% HCl  (Read 6446 times)

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#### taurean

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##### 5% HCl
« on: February 04, 2013, 10:07:03 AM »
How to make 5% HCl.  Can someone please make me understand the calculations?

Thanks,

#### discodermolide

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##### Re: 5% HCl
« Reply #1 on: February 04, 2013, 10:09:15 AM »
Say you have $100 and you give me 5% how many$ do I get?
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#### taurean

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##### Re: 5% HCl
« Reply #2 on: February 04, 2013, 10:12:22 AM »
Say you have $100 and you give me 5% how many$ do I get?

Yes, I understand that...but I think there are many things to consider like w/w, v/v and w/v and also the concentration of the commercially available HCl.  Am I wrong?  Don't know how to do the calculations based on all these factors.

#### Dan

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##### Re: 5% HCl
« Reply #3 on: February 04, 2013, 10:22:36 AM »
Don't know how to do the calculations based on all these factors.

Unless otherwise stated, 5% HCl means 5% w/w.

Try this one:

What masses of 37% HCl(aq) stock solution and water should be mixed to prepare 1 kg of 5% HCl(aq)?

Now this one:

What volumes of 37% HCl(aq) stock solution (density = 1.19 kg/L) and water should be mixed to prepare 1 L of 5% HCl(aq) (density = 1.02 kg/L)?
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#### discodermolide

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##### Re: 5% HCl
« Reply #4 on: February 04, 2013, 10:23:38 AM »
Well 5% is 5% be it w/w, v/v, or w/v. So where is the problem?
Concentrated HCl is a 37% solution I think w/w,(aldrich does not specify it) density of 1.2g/mL
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#### Babcock_Hall

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##### Re: 5% HCl
« Reply #5 on: February 04, 2013, 10:24:27 AM »
I think that your question correctly points out that the percent scale can be ambiguous.  I wrote up a two page (single-spaced) discussion of this one time, drawing upon several disciplines within chemistry for examples.  One interpretation of 5% HCl is that it is simply one part commercial HCl to 19 parts water (volume-to-volume), but that is not the only reasonable interpretation IMO.  Did the protocol from which you are working (or the question you are trying to answer) give any more information?  If not, I would go with Dan's interpretation.

#### AWK

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##### Re: 5% HCl
« Reply #6 on: February 04, 2013, 10:53:52 AM »
As a default we use w/w percentage for all solids and liquids (with one exception - ethanol solutions - default is v/v). Otherwise type of percentage should be clearly stated.
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#### Arkcon

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##### Re: 5% HCl
« Reply #7 on: February 04, 2013, 01:45:36 PM »
Or perhaps for the application, it isn't really significant.  Compare the molarity of an HCl solution at 5% w/w to 5% w/v.  Are they very far off?  If this acid solution just used to neutralize some reaction, or to wash an immiscible organic solution, maybe it doesn't really matter.
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#### confusedstud

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##### Re: 5% HCl
« Reply #8 on: February 06, 2013, 12:42:18 AM »
Don't know how to do the calculations based on all these factors.

Unless otherwise stated, 5% HCl means 5% w/w.

Try this one:

What masses of 37% HCl(aq) stock solution and water should be mixed to prepare 1 kg of 5% HCl(aq)?

Now this one:

What volumes of 37% HCl(aq) stock solution (density = 1.19 kg/L) and water should be mixed to prepare 1 L of 5% HCl(aq) (density = 1.02 kg/L)?

Hi Dan, I went to do the questions you posted and I'm not sure if I'm right

First one: Consider in terms of kg
So, 37kg+63kg+x=total mass
To find x, 37/(63+x)=5 so x=677kg
37+63+677=777 for the total mass to be 1kg, mass of 37% HCl water=100/777 and mass of water added=677/777=0.87kg

Is this correct?

For the second question I'm not very sure how to answer it. I tried to equate the total mass/ total volume to 1.19 but i didn't get it. Here is my working total mass=100kg total volume=63kg=63l so density=100kg/63L=1.59kg/L . How is this wrong?

I tried it again after some thinking and here are my new workings:

Consider in kg and using first answer,
100kg+677kg/xL=1.02kg/L so xL=762L

however it needs to be 1L instead of 762L so, 777/762=1.02kg. In order to get the respect amounts i split them into 100kg/762L+677kg/762L

so the new mass of solution is 0.131kg and mass of water is 0.89kg. For their initial densities, 0.131kg/V1=1.19kg/L so V1=0.11L and 0.89kg/V2=1kg/L so V2=0.89L

Is this correct? Also, is taught in high school chemistry? Because i did not learn about these stuff.
« Last Edit: February 06, 2013, 01:44:57 AM by confusedstud »

#### Dan

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##### Re: 5% HCl
« Reply #9 on: February 06, 2013, 03:05:24 AM »
First one: Consider in terms of kg
So, 37kg+63kg+x=total mass
To find x, 37/(63+x)=5 so x=677kg
37+63+677=777 for the total mass to be 1kg, mass of 37% HCl water=100/777 and mass of water added=677/777=0.87kg

Is this correct?

The numbers look OK, but I do not follow the logic of how you got there at all. Can you explain?

My approach would be:

Mass of HCl in 1000 g of 5% HCl(aq): 1000 x 0.05 = 50 g of HCl
Mass of 37% HCl containing 50 g HCl: 50/0.37 = 135 g
Final mass of solution is 1000 g, water required: 1000 - 135 = 865 g

So, 135 g of 37% HCl(aq) + 865 g of water gives 1000 g of 5% HCl(aq)
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#### confusedstud

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##### Re: 5% HCl
« Reply #10 on: February 06, 2013, 03:29:35 AM »
First one: Consider in terms of kg
So, 37kg+63kg+x=total mass
To find x, 37/(63+x)=5 so x=677kg
37+63+677=777 for the total mass to be 1kg, mass of 37% HCl water=100/777 and mass of water added=677/777=0.87kg

Is this correct?

The numbers look OK, but I do not follow the logic of how you got there at all. Can you explain?

My approach would be:

Mass of HCl in 1000 g of 5% HCl(aq): 1000 x 0.05 = 50 g of HCl
Mass of 37% HCl containing 50 g HCl: 50/0.37 = 135 g
Final mass of solution is 1000 g, water required: 1000 - 135 = 865 g

So, 135 g of 37% HCl(aq) + 865 g of water gives 1000 g of 5% HCl(aq)

Consider in kg,
37kg+63kg+xkg=total mass
percentage of HCl in final product=5%=37kg/100kg+x therefore x=640kg.
So total mass=100+640=740kg however as question states for it to be 1kg, 100/740=0.135kg (stock) and 640/740=0865kg

My logic is that in the 37% HCl and 63% water, i consider them to be 100kg in total. So from there to dilute the solution, i add x kg of water. So to find x, i use the final product's HCl percentage to solve for x. After this, i can get the total mass but since the question wants the total mass to be 1kg instead, i divide each variable by itself.

Will this approach be okay? And was this taught in high school because in Singapore where i just took my O levels this type of question weren't taught to us.

Thanks for the help

#### Dan

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##### Re: 5% HCl
« Reply #11 on: February 06, 2013, 06:22:26 AM »
Will this approach be okay?

Yes that's fine. We just visualise the problem differently so your logic was not immediately obvious to me, but I see what you're doing now and it is correct.
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#### confusedstud

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##### Re: 5% HCl
« Reply #12 on: February 06, 2013, 07:43:33 AM »
Hi thanks for the reply. But my workings in the second question is directly related from the first answer. However, if I was just given the second question I don't think I would be able to answer the question. What would be your approach to the second question?

#### Dan

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##### Re: 5% HCl
« Reply #13 on: February 06, 2013, 08:26:28 AM »
What would be your approach to the second question?

The same as the first one, just convert to mass first:

Quote
What volumes of 37% HCl(aq) stock solution (density = 1.19 kg/L) and water should be mixed to prepare 1 L of 5% HCl(aq) (density = 1.02 kg/L)?

1. What is the mass of 1 L of 5% HCl?
2. So what is the mass of HCl in 1 L of the solution?
3. What mass of the stock solution will you need to supply that HCl?
4. What mass of water will you need to mix with the stock?
5. Convert 3 and 4 to volume
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#### confusedstud

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##### Re: 5% HCl
« Reply #14 on: February 06, 2013, 09:02:16 AM »

Mass of 1L of 5% HCl=1.02kg
Mass of HCl in that solution=5/100 x 1.02kg=0.051kg
37/100 x mass of 37% HCl =0.051kg thus mass of solution=0.139kg
Mass of water=1.02-0.139=0.881kg

1.19kg/L=0.139kg/volume so volume=0.12L and the water is 0.88L

I think this is similar to my answer for the second question. But do you think my method is feasible because I simply used the total mass I found in the first equation for the 5% HCl solution.

Thanks for the help