# Chemistry Molarity Help regarding mixtures?

Indicate the concentration of each ion or molecule present in the solutions formed by mixing:
a) 20 ml of 0.100 M HCL and 10.0 of .500 M HCL
b) 15.0 mL of .300 M Na2SO4 and 10.0 mL of 0.200 M KCL

I'm in a A.P. Chemistry prep class and have no idea how to do these questions. I'm am most confused about mixing the solutions. It would be good if you explain the problem in detail

Answer:
a) Get the moles of each solution and divide it by the total volume. So,
moles of 1st HCl soln = 20.0 mL * 0.100 mol/L = 2.00mmol
moles of 2nd HCl soln = 10.0 mL * 0.500 mol/L = 5.00mmol

concentration of each ion or molecule = (2.00mmol + 5.00mmol) / (20.0mL + 10.0mL)
=0.233 mol/L or 0.233 M

b) Get the number of moles of each ion and then divide it by the total volume. So,
mole Na2SO4 = 15.0mL * 0.300 mol/L = 4.50mmol
mole KCl = 10.0mL * 0.200 mol/L = 50.0mmol

conc. of Na+ = (2 * 4.50mmol) / (15.0mL + 10.0mL)
= 0.360 mol/L or 0.360 M
note: there are 2 Na+ for every Na2SO4, hence value is times two

conc. of SO4^2- = 4.50mmol / (15.0mL + 10.0mL)
= 0.180 mol/L or 0.180 M

conc. of K+ = 50.0mmol / (15.0mL + 10.0mL)
= 2.00 mol/L or 2.00 M

conc. of Cl- = 50.0mmol / (15.0mL + 10.0mL)
= 2.00 mol/L or 2.00 M
Thank you
Thank you
Thank you
Thank you

For reminding me that I don't have to do this anymore!!!!!!!!!...
1) 3.1304 x 10^22 molecules of HCl (or ions of H+ and Cl- since the molecule breaks down to a one to one ratio of the two)

I'm not too sure that my math is right, I haven't done stoichiometry in a long time. What you do is you take the twenty ml and convert it into liters, which is .02 liters. Then you multiply that by .1 moles (1 mole per liter, hence Molar) to get .002 moles of HCL. Then by Avogadro's Law, you have 6.02x10^24 molecules (and in this case individual ions as well) of HCL (and it's components) in the first solution if it's one molar, which it isn't so you multiply it by .002 to get 1.204x10^22 molecules of HCl in the first solution.

You do basically the same thing for the second solution to get 3.01x10^22 molecules of HCL in the second solution. Since no reaction is taking place in the equation, you add the two numbers together to get the final result.

2) First, you'll have to balance the equation for this so that you'll have:

Na2SO4 + 2KCl -> 2NaCl + K2SO4

I'm not too confident in my abilities to do this one, sorry, but you should basically follow the same steps as for the first one, only you'll do it with the individual ions (2Na+, SO4-, 2K+, and 2Cl-)

Best of luck and sorry I couldn't give an exact answer for the second question.
i think for the second one, there's no reaction whatsoever. the equation silentuimide gave makes no sense

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