8
Jia Lab jie huo(10780081)
Result and Discussion
Result: In part 1, we try to mix a variety kind of solutions and find the results of
them. There is no reaction between NaCl and KNO3; there is white precipitation
and heat evolved when mixed NaCl with AgNO3; there is gas and heat evolved
when mixed NaCO3 with HCl; there is brown precipitation and heat evolved when
mixed BaCl2 with H2SO4; there is no result can be detected between NH3 and
H2SO4; there is no reaction between CuSO4 and Zn(NO3)2; there is white
precipitation between Na2CO3 and CaCl2; there is no reaction between CuSO4 and
NH4Cl; there is heat evolved between NaOH and HNO3; there is the red-brown
precipitation evolved between FeCl3 and NaOH; there is gas and heat evolved
between NaHCO3 solution and HCl. In part 2, , we first assumed that the CuSO4
and NaOH could react with each other because according to the solution law,
compound s containing OH- are insoluble unless the cation is one of Na, Li, K,
NH4. So Cu(OH)2 is unsoluble. In the actual experiment, we found that the mixture
of them had produced blue precipitation. In part 3, we tested the Ba(OH)2 and
Na2SO4 solutions and they both had conductivity. When they mixed together, the
solution had conductivity. Then, when tested the conductivity of Ba(OH)2 and
H2SO4, they both had conductivity. But when we mix them together, the
consequent solution had no conductivity.
Discussion: In this experiment, we went through smoothly and there were not
severe mistake happened. However, there may be three different kinds of errors
occurred during the experiment, first, when checked the results of the combination
of different solutions, we could not easily identify the result if the result is just evolving the
heat because some reaction may produced too few heat. Second, when in
the part three, we tested the conductivity of different solutions and also the
combination of them. However, we could not quite sure we measure the same
volume of Ba(OH)2 with Na2SO4 or H2SO4, so the result may somewhat
unpersuasive.
Conclusions:
These experiments show the results of the combination of different solutions
and test the rightness of the solution law. And we also know the conductivity of
different solutions and also that of their combination.
Post lab Questions:
1. Not yet find.
2. H2SO3(aq)H2O(l)+SO2(g)
3. a. K2S(aq) + CuSO4(aq) K2SO4(aq) + CuS(S)
( S2+ (aq)+ SO42- (aq) CuS(s) )
b. KOH(aq) + NH4Cl(aq) KCl(aq) + NH3(g) + H2O(l)
( OH-(aq)+ NH4+ (aq) NH3(g) + H2O(l) )
c.
(NH4)2SO4 + NaCl no reaction
d.
CoCl3(aq) + NaOH(aq) 3NaCl(aq) + CoOH3(S)
(Co3+ + OH- CoOH3(s))
e.
Na2CO3(aq) + 2HNO3(aq) 2NaNO3(aq) + H2O(l) + CO2(g)
(CO32- (aq) + 2H+(aq) H2O(l) + CO2(g))
4. a.
2Co(NO3 )3(aq) + 3Na2S(aq) Co2S3(s) + 6NaNO3(aq)
( 2Co3+(aq)+ 3S2- (aq) Co2S3(s) )
b.
H3PO4(aq) + 3KOH(aq) K3PO4(aq) + 3H2O(l)
( 3H+(aq) + 3OH-(aq) 3H2O(l) )
c.
2(NH4 )3PO4(aq) + 3CaCl2(aq) 6NH4Cl(aq) + Ca3(PO4)2 (s)
(2PO42-(aq) + 3Ca2+(aq) CaPO4(s))
d.
K2CO3(aq) + 2HCl(aq) 2KCl(aq) + H2O(l) + CO2(g)
(CO32-(aq) + 2H+ (aq) H2O(l) + CO2(g))
5. Both Ba(OH)2 and H2SO4 are ionic compounds so that they have conductivity
when in the solution state, but when
Ba(OH)2(aq) + H2SO4(aq) BaSO4(s) + 2H2O(l)
from the equation we know that one mole of Ba(OH)2 react with one mole of
H2SO4. So the resultants would be the precipitation of BaSO4 and H2O, and H2O
had no conductivity, so the mixture of the two does not have conductivity.