Exactly 1.00 × 10-3 mole of CaCl2, 0.100 mole of NaOH, and 0.100 mole of Na3T we

Question

Exactly 1.00 × 10-3 mole of CaCl2, 0.100 mole of NaOH, and 0.100 mole of Na3T were mixed and diluted to 1.00 liter. What was the concentration of Ca2+ in the resulting mixture? Consider Ca2+ + T3- CaT- where K = 1.48 x 108 Exactly 1.00 × 10-3 mole of CaCl2, 0.100 mole of NaOH, and 0.100 mole of Na3T were mixed and diluted to 1.00 liter. What was the concentration of Ca2+ in the resulting mixture? Consider Ca2+ + T3- CaT- where K = 1.48 x 108 Exactly 1.00 × 10-3 mole of CaCl2, 0.100 mole of NaOH, and 0.100 mole of Na3T were mixed and diluted to 1.00 liter. What was the concentration of Ca2+ in the resulting mixture? Consider Ca2+ + T3- CaT- where K = 1.48 x 108

Explanation / Answer

Since CaCl2 is completely soluble in water, initial con. of Ca2+, [Ca2+] = 1.00x10-3 mol / 1.00L = 1.00x10-3 M

initial concentration of T3-, [T3-] = 0.100 mol/ 1.00L = 0.100 M

The balanced chemical equation is:

-------------- Ca2+(aq)   + T3-(aq)   <-----> CaT-(aq) : K = 1.48x108

Init.con(M):1.00x10-3, ---- 0.100 ------------ 0

eqm.con(M):(1.00x10-3 - y), (0.100 - y) ---- y

K = 1.48x108 = [CaT-(aq)] / [Ca2+(aq)]x[T3-(aq)]

=> 1.48x108 = y M / (1.00x10-3 - y) M x (0.100 - y) M

=> 1.48x108 = y / (1.00x10-4 - 1.00x10-3y - 0.100y + y2)

=> 1.48x104 -1.4948x107y - y +1.48x108y2 = 0

On solving the above equation we get, y = 0.10198 M

y = 0.0009999999317499356 M

Hence equilibrium cncentration of Ca2+, [ Ca2+] = (1.00x10-3 - y) = (1.00x10-3 - 0.0009999999317499356)M

= 6.82x10-11 M (answer)

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