Question

What is the concentration of nitrate ions if 45 mL of 0.25 M Sr(NO3)2 are diluted to 68 mL? Factor label please!!

Answer 1

Sr(NO₃)₂ is an ionic solid which dissociates as

Sr(NO₃)₂ Sr²⁺ + 2 NO₃^-

initial concentration of Sr(NO₃)₂ = 0.25 M

initial concentration of NO₃^- = 2 * initial concentration of Sr(NO₃)₂

initial concentration of NO₃^- = 2 * 0.25 M

initial concentration of NO₃^- = 0.50 M

According to dilution law, M1 * V1 = M2 * V2

where M1 = molarity before dilution

V1 = volume before dilution

M2 = molarity after dilution

V2 = volume after dilution

for NO₃^- , M1 = 0.50 M , V1 = 45 mL , V2 = 68 mL

M2 = (M1 * V1) / V2

M2 = (0.50 M * 45 mL) / 68 mL

M2 = 0.50 M * (45 mL / 68 mL)

M2 = 0.50 M * (0.662)

M2 = 0.33 M

Concentration of nitrate ions after dilution = 0.33 M