# Indicate the concentration of each ion present in the solution formed by mixing

Indicate the concentration of each ion present in the solution
formed by mixing the following. (Assume that the volumes are
additive.) (a) 30 mL of 0.100 M HCl and 10.0 mL of 0.520 M HCl H+ _____M Cl-______ M (b) 15.0 mL of 0.294 M Na2SO4 and 28.2 mL of 0.200 M KCl Na+______ M K+______ M SO42-______ M Cl-______ M (c) 3.50 g of NaCl in 59.3 mL of 0.439 M CaCl2 solution Na+______ M Ca2+______ M Cl-______ M

a.)
Mole HCL = 30 x 0.100 / 1000 = 0.003
Mole HCl = 10 x 0.520 / 1000 = 0.0052
Total mole = 0.003 + 0.0052 = 0.0082
Total volume = 30 + 10 = 50 mL = 0.040 L
HCl is a strong acid >> H+ + Cl-[H+] = [Cl-] = 0.0082 / 0.040 = 0.205 M
b.)
Mole Na2SO4 = 15 x 0.294 / 1000 = 0.00441
Na2SO4 >> 2Na+ + SO42-
Mole Na+ = 2 x 0.00441 = 0.00882
Mole SO42- = 0.00441
Mole KCl = 28.2 x 0.200 /1000 = 0.00564
KCl >> K+ + Cl-
Mole K+ = mole Cl- = 0.00564
Total volume = 15 + 28.2 = 43.2 mL = 0.0432 L[Na+] = 0.00882 /0.0432 = 0.204 M
[SO42-] = 0.00441 / 0.0432 = 0.102 M
[K+] = [Cl-] = 0.00564 / 0.0432 = 0.131 M
c.)
c)
Number of moles of NaCl = 3.5/(35.5+23)=3.5/58.5 = 5.98
*10^-2
Number of moles of Na+ = 5.98 *10^-2 moles = 0.0598 moles
Total number of moles of Cl- = 2*0.439 * 0.0593 + 5.98*10^-2 =
2*0.0260327+0.0598 = 0.112 moles
Molarity of Na+ = 0.0598*1000/59.3 = 1.008 M
Molarity of Ca+2 = 0.439 M
Molarity of Cl- = 0.112*1000/59.3 = 1.888 M

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