a friend got a 30% hcl solution and has no ph meter. his friend allready researched in the internet the ph of the concentrated solution, finding this:

Assuming there are 300g HCl and 700gH2O you have 8.23 moles of HCL
HCl has a density of 1.15 Kg/L at 30% so 1000g/1.15Kg/L=869.57mL solution
Molarity= moles/Liters of solution.
8.23mol/.86957L = 9.46M
pH=-log[H+] and [H+] is the same as the [HCl] because is a strong acid.
-log(9.46)=-.98
30% HCl has a pH of -.98

the other question he has, and could not find an answer is:

how much hcl30% does he need to turn 1L of a water solution with a ph of 4,7 to a ph of 2?

also swim notided th in my friend bottle there is written: Cloridric Acid S/HF 30%.
what is the meaning of S/HF?


thank you!

I'm by no means an expert...And its late..But have some experience in the field. There may be a more efficient/accurate method for approaching this problem, but i can help give a rough general estimate.

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The molarity you calculated is slightly off:
from 1kg of 30% HCL Soltn; .3 = HCL, .7 = H2O.
.7kg H2O = .7 L (density is 1kg/L)
.3kg HCL = 0.261 L (density is 1.15kg/L)
so, total volume is approx: .961 L
moles of HCL calculated was correct though, 8.23 mol (Molar mass of HCL = 36.45g/mol)

-30% HCL Soltn. is 8.56 M (8.23mol/.961 L)

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To calculate necessary amount, i determined the moles required for the pH change, ignoring the additional volume added to achieve this concentration.

(10^-2)-(10^-4.3)= 0.00994988127 M [H3O+] difference, in 1 L of solution moles HCL is same value.

HCL, being a strong acidic electrolyte, will fully dissociate in a 1:1 stoichiometric ratio. You can use the molarity determined previously to determine volume needed to dissociate this number of moles of HCL.

8.56 M = (.00995 mol/ X Liters)
(.00995 mol/8.56 M) = X Liters
0.00116238317 = X Liters
(0.00116238317 L)(1000 ml/ 1 L)= 1.16238317 ml (30% HCL solution)

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so.. to approximate final pH, and check previous calculations:

(10^-4.3 mol)+(0.00994988127 mol)= 0.00999999999 mol HCL
(0.00999999999 mol HCL / 1.00116238317 L)= 0.00998838965 M [H3O+]

-log[0.00998838965] = 2.00050452404 pH (approximate final pH)

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Like i said there may be a more accurate/efficient way of going about the same problem, but hopefully this gives a good enough rough approximation. I hope that someone with more expertise in the field can double-check my logic. I'm sorry i don't know what "Cloridric Acid S/HF 30%" refers to.

If you don't have a pH meter, you could try cabbage juice or litmus paper (pH test strips-if you can find it).
You can buy a pH test kit at a hydroponic store, but those don't read below like pH 5.

GOOD LUCK!

I am certainly not a chemical expert, so this is how I would reason, coarsely using some rules of thumb:

If the bottle of HCl has a pH of -1, then diluting that tenfold would give a solution with pH 0. Diluting that tenfold gives pH 1, and again diluting tenfold gives pH 2.

So, taking 1 milliliter from the pH -1 30% HCl bottle and adding it to 1 liter of pH7 water gives about a liter of pH 2 0,03% HCl solution.

Now if I take 1 milliliter from the pH 2 solution and again add it to 1 liter of pH7 water, I get about a liter of pH 5 HCl solution, almost pH 4.7. If I added a milliliter more, I might even get quite close to 4.7.

So what would happen if I took a liter of this last solution, with a pH of around 4.7, and added a milliliter of pH -1 30% HCl? About the same as taking a fresh liter of water and adding 1.002 milliliter of pH -1 30% HCl.

In other words, the starting pH of 4.7 is almost negligible when acidifying to pH 2.

Note that this reasoning assumes the ideal situation involving only water and HCl. In the real world, your pH 4.7 might contain other ions, this is called a buffer solution. A buffer can diminish the effect of adding the 30% HCl a bit, but not by an order of magnitude.

As I understand it, 30% HCl is usually calculated as weight per volume (what with HCl being a gas and all).

This means each litre of solution contains 300 g of HCl.

MW (HCl) = 36.453; 300/36.453 = 8.229775327 M

So 30% HCl is about 8.23 molar as originally assumed; I'll spare myself the fun of the pH calculations

Cloridric Acid seems to be an anglification of the French, "acide chloridrique" but I've not the foggiest idea what S/HF refers to. It's certainly not important (IMO).

I didn't think it was possible for something to have a negative ph.

Sure...The Alien uses it for blood.