I'm having trouble tackling some chem questions...please help!!!
Kaplan Integration B #58:
As an aqueous sulfuric acid solution is 39.2% H2SO4 by mass and has a specific gravity of 1.25. How many milliliters of this solution are required to make 100 mL of a 0.2M sulfuric acid solution?
The answer shows that you divide the mass of H2SO4 by the percentage of 39.2%. This doesn't make sense to me. Answer is 4mL
Solution weight = solute weight + solvent weight
Density: 1.25kg solution/ 1L solution
Let's say you have 1L of this solution. Then due to density (a static value), this much solution should have 1.25kg solution weight. Of this, 1.25kg x .4 = 0.5kg is solute (H2SO4) and 1.25kg x .6 = .75kg is solvent (could be water). Since 0.5kg is the weight of H2SO4 (MW = ~100g/mol). The H2SO4 mole # in 1L of the solution is 500g / (100g/mol) = 5 mol H2SO4 in 1L solution. This also conveniently tells us that the solution is 5M solution. Does the concentration of 1L of the solution different from X liters of the same solution? No.
The next step seems easy at first but conceptually quite difficult. You have 5M H2SO4 solution. What do you do to make 100 mL of 0.2M H2SO4 from 5M solution? In a real experient situation, since you cannot take out solute alone from the solution, you would dilute 5M solution
by adding
solvent until you reach .2M and take out a sample of 100mL from the new solution. There you have it: 100mL of .2M solution.
However, the problem asks you to figure out
how much of 5M solution (solute+solvent) is required to make 100mL of 0.2M solution,
not how much solvent to add or subtract. To emulate the situation here in a real experiment, you would take some portion
of 5M
solution and evaporate all the solvent and add 100mL to the dry solute then you get .2M solution. What they are asking you to do is to calculate exactly how much of that 5M solution will have to be separated from the solution so that when evaporated and added 100mL of fresh solvent, you would end up with .2M solution. First you want to figure out how much solute have to be present in the separated solution before isolating the dry solute: 1L x .2M = .02 moles is what we want. Now what volume of 5M solution would have exactly .02 moles of solute? .02 mole/ X liters = 5M. You do the math and you end up with 4mL.
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Achiever test 2 #41:
How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?
Is there a faster/easier way to calculate this problem than what they give you in the solution? Answer is 30mL
HCl + KOH -> H2O + KCl + H+ +Cl-
Init .5a .2x.05 = .001 0 0 0
Change: -.001 -.001 .001 10^(-2) 10^(-2)
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Final: .5a-.001 0 .001 10^(-2) 10^(-2)
H+ from HCl completely dissociates in pure water and reacts completely with KOH. Remember that HCl and KOH react at 1:1 ratio (mole# ratio). So if you have access HCl (HCl mol# > KOH mol#), then you will have extra H+ in the product side. Of 0.5x mol of H+, .001 mol is used up by neutralization reaction with OH- from KOH, making .001mol new H2O. Thus, 0.5a-.001 H+ should be how much H+ you end up with. This is given as 10^(-2) M by pH.
0.5a-.001 = 10^(-2) x (a+.02), solving this gives you a = .03 liters
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Kaplan test 2 #53. What is the molality of 1L aqueous solution containing 31 percent HCl by mass?
I'm getting my concepts mixed up with this one. If you assume a 100 gram sample than that gives you 70 grams of H2O for the solvent, which is 0.07kg H2O. What confuses me is what do you do with the 1.0L solution? Since we are dealing with molality do we just ignore the 1.0L solution.. is it just extra info? Answer is 12 m.
Think of it this way: let's say you have 20 liters of some solution with 1M or 1m. Does the concentration change if you took out 2 liters of this solution or .354 liters of this solution? If you took out or added 2 liters of
solvent then the concentration
would change. Same analogy goes here. Let's say you have 100g of aqueous solution having 5g of HCl dissolved in it. Assume this solution is homogenous. If you isolated 10g of this solution in a separate container, isn't it logical to expect .5g of HCl to be dissolved in it? This intuition is correct because 5% HCl solution is 5% HCl dissolved in any volume of the solution. The concentration does not change becuse you took 100mL portion of the solution or 1mL portion of the solution. 1L in this problem is totally irrelevant. In fact you caught the distractor correctly cause' it would have been useful if they asked you about molarity.
To solve this problem let's assume u have 100g of solution.
3/10 x 100g = 30g HCl ~ 0.85 mol HCl
7/10 x 100g = 70g H2O = 0.07 kg solvent
m = .85mol /.07kg = 12.1
Oh boy... I can't believe that I spent over an hour writing this... but I hope someone looks at it, appreciate my effort
! It was a good review for myself! Good luck on the test!
