1) If ZnCO3 (solid) is insoluble in water, but dissolves in excess 6.0 molar HCl (aqueous), how would pull out the proper equation that would support it?
2) How would you go about finding how many mL of 2.00 molar NaOH (Aqueous) solution are required to neutralize 1.858 g of KHC8H4O4 (it's formula weight being 204.22)? Any other example would do.
NaOH(aq) + KHC8H4O4 --> NaKC8H4O4(aq) + H2O(l)
3) Potassium nitrate, KNO3, has a formula weight of 101.10. How would I find the molar concentration of a solution prepared by dissolving 7.58 grams of potassium nitrate in enough water to prepare 250 mL of the solution?© BrainMass Inc. brainmass.com October 24, 2018, 6:12 pm ad1c9bdddf
1. ZnCO3 doesnot dissolve in water, however in the presence on a strong acid we can have a displacement reaction: meaning the cation of the first compound will join with the anion of the second and vice versa such that:
ZnCO3 + 2HCL ------> ZnCL2 + H2CO3
since Zn has a charge of +2 and cl has a charge of -1, we need 2 cl's for every Zn and so writing the equation, we know that the ration of HCL is twice that of ZnCO3. On the other hand now we have 2H's for ever CO3. This also makes ...
The solution goes through these questions of solubility, molarity and neutralization one by one, explaining the theory and applying the calculations for clean, clear answers.
Molar solubility product of barium fluoride
Please help with the following problems. Provide step by step calculations for each.
5.[a] The molar solubility of barium fluoride, BaF2, is 7.5 x 10-3 M. Calculate the solubility product
[b]Calculate the molar solubility of lead fluoride, PbF2 in water, Ksp = 2.7 x 10-8
6.Calculate the molar solubility of bismuth sulfide, Bi2S3 (Ksp = 1.6 x 10-72) in (a) pure water, and
(b) 0.05 M cadmium sulfide, CdS