How would your calculated value of K_sp be affected by errors introduced from th

Question

How would your calculated value of K_sp be affected by errors introduced from the following affected? Be specific in your answer. State K_sp would too large, too small, or not affected and explain why that would have that effect on K_sp. a. The precipitate of Ca(IO_3)_2 when obtained originally. is not washed with distilled water. b The Na_2S_2O_3 was not standardized. so its nominal molarity was used instead of a standardized molarity Although the KIO_3 used to standardize the Na_2S_2O_3 is labeled as 0.0100 M. its actual concentration is somewhat higher than 0.0100 M. d Extraneous water Is introduced from wet filter paper or wet funnels, or by washing the precipitate during the final filtration of Ca(IO_3)_2.

Explanation / Answer

a) In the initial step of this experiment, you need to precipitate Ca(IO3)2 by mixing solutions of KIO3 and Ca(NO3)2.The white precipitate of Ca(IO3)2 then gets separated.

But ,if the precipitate is not washed properly with distilled water it may have traces of ions of the reactants-Ca2+,IO3-, that may give an erroneous value of ksp as Ksp=[Ca2+][IO3-]2

The value of ksp will be too large.

for the solubility of Ca(IO3)2 as per the equation Ca(IO3)2(s)<------->Ca2+(aq) +2IO3-(aq)

b)The titration of IO3- is done using standardised Na2S2O3 generally.

The reaction that takes place is-

2IO3- +10I- + 12H+ ------>6I2 +6H2O

I2+2S2O32- --->2I- +S4O62-

I2 is (blue in starch),which turns colorless at the end point

So,if Na2S2O3 is not standardized, the concentration of S2O32- =[S2O32-]=1/2[I2]

Also ,6[I2]=2[IO3-] or 3[I2]=[IO3-]

or,[I2]=1/3[IO3-]

So,finally,[S2O32-]=1/2[I2]=1/6[IO3-] which gives 6[S2O32-]=[IO3-] =S

ksp=Ksp=[Ca2+][IO3-]2

Also ,if solubility of IO3-=S then [Ca2+]=1/2*S

ksp=S*(1/2*S)2=1/4S3=1/4 *6[S2O32-]=3/2 [S2O32-]

So,for higher [S2O32-],ksp will be high

c)8IO3- +5S2O32- +2OH- --->4I2 +10SO42- +H2O

Hence ,M(IO3-) *V(IO3-)/M(S2O32- ) *V(S2O32- )=8/5

M=molarity,V=volume

So, M(S2O32- ) =(5/8)M(IO3-) *V(IO3-)/V(S2O32- )

if M(IO3-) has higher value M(S2O32- ) will be higher ,so high ksp as per the eqn in part b)

d)No effect on ksp as ksp solely depends on the concentration of Ca2+ and IO3- and water generally do not have these ions.

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