indicate the concent. of each ion present in soln' formed by mixing...?

Question

16.0mL of 0.130M HCl and 12.0mL of 0.600M HCl.
B) 18.0mL of .200M Na2SO4 and 15.0mL of .150M KCl.
C) 2.38g of NaCl in 50.0mL of .400M CaCl2 soln'

Answer 1

A)

find # of moles of each substance.
#1 HCl = (.13M * .016L) = .00208moles
#2 HCl = (.6M * .012L) = .0072 moles
Total moles = .00928moles
Total volume = .028L
Concentration = .00928 moles / .028L = .33M

B)

find number of moles
Na2SO4 = (.2M * .018L) = .0036moles
KCl = (.15M * .015L) = .00225 moles
Total volume = .033L
Na2SO4 concentration = .0036moles / .033L = .109M
KCl concentration = .00225moles / .033L = .068M

C)
find number of moles
Molecular weight of NaCl = 58 g / mole
NaCl = 2.38g / 58 g/mole = .04moles
CaCl2 = (.4M * .05L) = .02 moles
Since adding NaCl will not change the volume adversely, the concentration of CaCl2 will not change.
NaCl = .04moles / .05 L = .8M
CaCl2 = .4M

Hope this answers your question.

Answer 2

16mL of 0.130M HCl has 0.016*0.130 moles of HCl.
12.0mL of 0.6M HCl has 0.012*0.6 moles of HCl.

Mix them together and you get a total of
0.016*0.130 + 0.012*0.6 moles HCl
which are now in a 12 + 16 = 28.0mL solution.

The concentration is then

(0.016*0.130 + 0.012*0.6) / 0.028 moles/L

If you assume 100% dissociation, that is the same as the ionic concentration of H+ and Cl-

Do the others the same way.

Answer 3

The answer to everything is 2.