Determining Acid Concentration for Chemistry Students

[saranghae]

Hi, I got this question for my chemistry class, and I simply don't know where to start. Do I have to find out the volume of HCl first?

The label on a bottle of hydrochloric acid says the solution is 3.7% hydrogen chloride. You wish to the determine the actual concentration. You fill a burette with the acid, but before you do the titration, you would like to know how much acid you need to add to the 10 mL of 1.0 mol/L sodium hydroxide to turn the blue bromthymol blue solution green. You can assume the density of the dilute acid is essentially that of water.

 

[mrjeffy321]

While neutralizing a Sodium Hydroxide solution with HCl the following reaction occurs,
NaOH + HCl --> NaCl + H2O,
For every 1 moles of NaOH present in solution, 1 mole of HCl is required to neutralize it and produce 1 mole of NaCl and H2O.

First, I would find the number of moles of NaOH in 10 mL of a 1 Molar (mol/L) solution, this would also be the same number of moles of HCl needed to neutralize it.
Now that you know the number of moles of HCl required, now you must calculate the volume of HCl solution needed. Your acid solution's concentration is given as a % mass of HCl (the rest being water), I think it would be easier to deal with this as a Molar concentration.

To convert x % mass to Molar, assume you have 100 grams of the solution, which means you x grams of HCl, which we know is (x / molar weight of HCl) number of moles. That number of moles is dissolved in the remaining weight of water (100 - x) grams. If we assume the density of this acid is 1 g/mL, then we can easily get the volume of the solvent. Once you have both these number, calculate the Molar concentration.

Back to the problem at hand. Now that you know the molar concentration of the acid (moles / Liter) and you know the number of moles you need, now it just is a matter of solving for the volume.
(Molar concentration = moles / volume) --> (Molar concentration * volume = moles)

 

[saranghae]

Thank you so much!
When you say molar weight of HCl, do you mean 36.5 g/mol?

 

[mrjeffy321]

yes,
(1.01 g/mol of H) + (35.45 g/mol Cl) = 36.46 g/mol HCl

 

[nguyenh]

Hi all,
I want to ask you about some problem (help my friend):
I have:
1)12M HCl, MG: 36,46g, Density:1,16g/cm3 (37% HCl=12M)

I have already calculation: 12M of HCl=12mol/lx 36,46g/mol=437,52g/l.

So, How much (ml) of 37% must need to prepare a 12M HCl solution? (How to calculate?)

1)6NaOH, MG: 40g, Density:1,353g/cm3 (19,62% HCl=12M)

I have already calculation: 6M of NaOH=6mol/lx 40g/mol=240g/l.

So, how much (g) solid NaOH nee prepare for 6M NaOH?

And one the question: How can calculation % (37%and 19,62%) from density? I have this result from my friend.

Many thanks


Please help me as soon as.

 

[sdekivit]

Hi all,
I want to ask you about some problem (help my friend):
I have:
1)12M HCl, MG: 36,46g, Density:1,16g/cm3 (37% HCl=12M)


1)6NaOH, MG: 40g, Density:1,353g/cm3 (19,62% HCl=12M)

So, how much (g) solid NaOH nee prepare for 6M NaOH?

And one the question: How can calculation % (37%and 19,62%) from density?

you know that the % is a m/m % ? --> for 1) 37 % HCl = 37 g HCl / 100 g solution

now we need to go from the unit g HCl/g to mol/L. You can convert g HCl to mol HCl with M_{W} and g solution can be converted to L solution using the density.

So for the first one: 37% HCl = \frac {(37/36.46)} {(100/1.16) \cdot 10^{-3}} = 12 M

Now you can apply the same kind of calcualtion to get to know the mass of NaOH needed to prepare a 6 M NaOH solution with a volume for example 100 mL