Question

A. 6.7 g of K3PO4 are dissolved in 500 mL of water. What is the concentration (in normality) of PO4^3- ions in solution?

B. Calculate the formal concentration (in molarity) of a KCl solution if 18.6 g of KCl are dissolved in 500 mL of water?

C. What NaCl concentration results when 234 mL of a 0.830 M NaCl solution is mixed with 502 mL of a 0.370 M NaCl solution?

Answer 1

a)

Normality of PO4-3 ions

N = n*M

M = mol/L and n = number of equivalents

PO4-3 need 3 H+ so it has n = 3 equivalents

for molarity:

V = 500 ml = 0.5 L

m = 6.7 g of K3PO4

MW of K3PO4 = 212.27 g/mol

calcualte mol of K3PO4

mol = masS/MW = 6.7/212.27 = 0.0315635 mol of K3PO4

since 1 mol of K3PO4 will give 1 mol of PO4-3 then

0.0315635 mol of K3PO4 = 0.0315635 mol of PO4-3

then

M = mol/L = 0.0315635/(0.5) = 0.063127 M

therefore

N = 3*0.063127 = 0.189381

b)

M = mol /L

mol = masS/MW

MW KCl = 74.5513

mol = 18.6/74.5513 = 0.24949mol of KCl

M = 0.24949/(0.5) = 0.49898 M of KCl

Formality = M (in this special case)

C)

V1 = 234 ml

M1 = 0.83

V2 = 502 ml

M2 = 0.37

Total Volume = V1+V2 = 234+502 = 736 ml = 0.736 L

mol of NaCl = M1V1 +M2V2 = 0.234*0.83 + .502*0.37 = 0.37996 mol of NaCl

Mfinal = mol final / L final = 0.37996/(0.736) = 0.51625 M of NaCl