The Chemistry Help Blog

icerosephoenix asked:

Calculate the value of a silver dollar using the following information: (a) 20 silver dollars weight one pound, (b) each silver dollar is 90.0% silver and (c) pure silver is selling for $29.53/oz

Well,

There are 16 ounces in a pound, so $29.53 * 16 ounces/pound = $472.48/lb
1 silver dollar * (1 lb / 20 silver dollars) = .05 lbs
.05 * .9 parts silver/silver dollar = .045 lbs of silver
.045 * $472.48/lb = $21.26 / 1 silver dollar

tramthepoline asked: A solution of H2SO4 was found in a laboratory without a concentration label. Two steps were taken to analyze it. First, a 25 mL sample was diluted to a volume of 500 mL. Then, a 35 mL portion of the diluted solution was tested and found to require 42.16 mL of 0.1050 M NaOH to completely neutralize it. Calculate the molarity of the original, undiluted H2SO4 solution.

First off - I love acid/base problems, and this one’s especially nice because it uses a strong acid (H2SO4) and a strong base (NaOH). Strong means that there’s no fiddling around with Ka or Kb, just glorious, glorious stoichiometry. 

First off, we need the equation.

H2SO4 + NaOH = H2O + Na2SO4 (which will dissolve in solution and generally is an unimportant quantity!) is the unbalanced equation.

H2SO4 + 2NaOH = 2H2O + Na2SO4 is the properly balanced equation.

So, let’s take stock of what we know. 42.16 mL of .1050 NaOH is enough to figure out the number of moles of OH-, which will be equal to the number of moles of H+. After all, Molarity x Liters of Solution = Moles, or M x L = m. 

There are 1000 milliliters in a liter, so (42.16/1000) = .04216 moles of OH-. 

This is when you run into a potential zone of error, and it’s because H2SO4 has two H+ ions per molecule, and NaOH just has one OH- ion. So while moles of H+ are definitely going to be equal to moles of OH-, moles of H2SO4 are not going to be equal to moles of NaOH.

There are two ways you can solve for moles of H2SO4. You can:

a. Know that there are .04216 moles of H+, and that there are twice as many H+ ions as there are H2SO4 molecules, and so deduce that H+ moles / 2 = moles of H2SO4, or .04216/2 = .02108 moles

b. Use mole ratios in the original equation. If you look, there are 2 NaOH molecules to every one H2SO4 molecule. The logic follows - moles NaOH / 2 = moles of H2SO4, or .04216/3 = .02108 moles of H2SO4. 

But, as you know, we’re not done here. First, you need to figure out the molarity of the diluted H2SO4 solution. By now in Chemistry you should be aware that M x L = M x L, or Molarity x Liters = Molarity x Liters, because no matter how you dilute it, you still have the same number of moles. 

So, first, we divide .02108 moles by .5 liters to find the diluted molarity.

.02108/.5 = .04216 M (isn’t that number looking familiar? don’t worry, it’s just a coincidence that there’s a 2-1 mole ratio and a half-liter solution)

Once you know the diluted solution is .04216 M, you can reckon that-

.04216 M (.5 L) = X M (.025 L)

X = .8432 M H2SO4

You may notice that you probably could have skipped the step of finding out the molarity at the diluted state, and that’s true, but it’s always good to understand your equations.

Chemistry has started! Well, it’s started for me. I’m our resident Chem teacher’s faculty assistant. Today I set up some computers with Vernier Lab Equipment and set up some burets for a titration lab. Have any of you started on labs?