Commercial concentrated HCl:

Specific gravity = 1.19

1.19g of HCl in 1ml of water

Assay= 37.4%

37.4ml of HCl in 100ml of water

i.e. 37.4 x 1.19 = 44.506g of HCl in 100ml of water

Formula weight = 36.46

1M = 36.46 g HCl in 1000ml of water

So if 44.506g of HCl is present in 100ml of water

Or 445.06g of HCl is present in 1000ml of water

Molarity of that solution is 445.06 / 36.46 = 12.2

Thus molarity of concentrated HCl is 12.2 M

The pH is 1. HCl doesn't exists in 100% solutions, 37% is the maximum at room temperature and pressure. if that is what you have dilute it 12 times!

Hello Nada,

There is actually no 100% solution of HCl, 38% is best you can get under normal storage conditions. The molarity of 38% HCl is 12.39M. To calculate the pH you can use the equation: pH= -log[H3O+]. In case of HCl (very strong acid) you can assume that it is 100% ionised in H2O. Therefore, pH= -log (molarity).

Regards,

Thomas

pH-values for very strong acids are easy to calculate, because the concentration of H+ (or H3O+) is identical to the concentration of HCl in the solution. Thus, with pH = -log concentration of H+, the 1 M solution of HCl will give the pH of -log 1 or 0 (Zero), because 10 ^0 = 1. Not complicated.

The pH explained by Dr. Wostemeyer is very correct. To add to it, there is no 100% HCl available. So to calculate the molarity of HCl that you get from Market you need to check its specific gravity and Assay.

Specific gravity indicates the grams of HCl in 1 g of water (here) or any solution. 1g of water is equivalent to 1ml of water. So its is grams of HCl in per ml of water. Say if specific gravity is 1.19 then it indicates that 1.19g of HCl is present in 1ml of water.

The assay determines the amount of HCl in 100ml of water. So if the assay is 37.4% indicates 37.4 ml of HCl solution in 100ml of water.

With the examples quoted above 37.4 ml of HCl solution should contain 44.506g of HCl.

i.e now 44.506 g of HCl is present in 100ml of water.

Since 1M HCl solution is gram mole of HCl in 1L water it would be 36.46g of HCl in 1L of solultion.

in the example quoted 445.06g of HCl is present in 1L of water. Thus the molarity of that solution would be 12.206.

So approximately the concentrated HCl that is available is generally is 12M. and to make it 1M need to dilute it atleast 12 times.

Commercial concentrated HCl:

Specific gravity = 1.19

1.19g of HCl in 1ml of water

Assay= 37.4%

37.4ml of HCl in 100ml of water

i.e. 37.4 x 1.19 = 44.506g of HCl in 100ml of water

Formula weight = 36.46

1M = 36.46 g HCl in 1000ml of water

So if 44.506g of HCl is present in 100ml of water

Or 445.06g of HCl is present in 1000ml of water

Molarity of that solution is 445.06 / 36.46 = 12.2

Thus molarity of concentrated HCl is 12.2 M

Thank you all for the clear explanations. I just want to summarize your explanations by using this simple equation.

M = (10 * d*%C)/Mwt …….................….eq 1

Where:

M = Molarity, mol/L

D = density or specific gravity, g/ml

%C = percent concentration (mass % or Volume %)

 Mwt = Molecular weight, g/mol

Using dilution equation:

 M1 V1=M2 V2……...............................eq2

Substituting the value of M from eq1 into eq2 gives,

 %C1 V1=%C2 V2…............................eq3

These three equations will help us to calculate the Molarity or %C  of Dilute solution we want to prepare from a concentrated solution or how much of the concentrated solution we need to prepare the desired amount of dilute solution.

Thanks everyone, your explanation is useful.

just to your knowledge, 100% HCl is not behaving as an acid, it is just HCl gas, once it touch water molecules either from atmosphere or solution then H3O+ is generated which is acting as an acid source. With regard to calculation, it is clearly described above.

The question is asking for pH.  Molarity is 1 M as given in the question.

A 100% solution will have 100 g HCl in 100 mL of water.  If you want 1M HCl, (36.46 g /L) then take 36.46 mL of your 100% solution and dilute it with distilled water to one liter.

Shouldn't pH of 1 M HCl = pKa ?

@David: It's more complicated: pH equals pKa only at the equivalence point of buffer solutions. In that case the Henderson-Hasselbach equation applies:

pH = pKs - log (conc of HA/ conc of A-)

If the concentration of undissociated acid and the concentrations of the corresponding anion are the same, log (HA/A) is log 1 or zero.

For the strong acid HCl you cannot reach equimolar relationships between the protonated form and the anion. At 1 M, HCl is essentially completely dissociated. The pKs is very small (around 10-7) and thus far away from the pH, which is zero (-log 1).

M1V1=M2V2 may be used for any dilution. Where M1 stands for molarity of stock, V1 is volume of stock required to make the desired solution, M2 is the molarity of the future solution and V2 is the how much volume to be prepared.

to grasp the concept of molarity is the most critical in stoichemistry calculation. In this case, we need to find the weigh of HCl in g that Anish showed in his answer. Specific gravity = g/ml. 37.4% of HCl solution means it contained 37.4 ml of HCl/100 ml solution. Remember percent is not a unitless. The weigh of HCl in 100 ml of 37.4% solution is 37.4 ml HCl X 1.19 g/ml = 44.506 g HCl/100 ml solution.--> 445.06 g/l. Molarity = mole/l. The usage of mole concept is related to Avogadro's number, i.e. 1 mole = 6.022×1023 mol-1 contained this quantity or number of atoms, molecules or substances. e.g 1 mole of Na OH will react completely with 1 mole HCl to produce NaCl and H2O.

Let us suppose that you have available 37 wt% hydrochloric acid of some commercial brand (reagent grade). Unless stated otherwise, the concentration indicated in the label, should be assumed as given in wt%. The corresponding (approx.) density should be available from the label, but we can also take it as 1.184 g·cm-3 (cf. e.g. Perry's Chemical Engineers' Handbook). The molecular mass of HCl is 36.46 g·mol−1. From this data the molarity of the conc. sol. is readily calculated. The intended dilute solution (1 M) should be prepared by considering the appropriate (volumetric) dilution factor.