A solution contains one or more of the following ions: Hg2 2+,Ba2+ ,Fe2+ and . When potassium chloride is added to the solution, a precipitate forms. The precipitate is filtered off and potassium sulfate is added to the remaining solution, producing no precipitate. When potassium carbonate is added to the remaining solution, a precipitate forms.Fe2+ and Hg2 2+ were present in the original solution.
Write net ionic equations for the formation of each of the precipitates observed.
A is a solution of tetraoxosulphate (vi) acid .B is a solution containing 1.4g of potassium hydroxide per 250cm cube (a) put A into the burette and titrate with 25cm cube portion of B using methy1orange or screened methy1orange as indicator . Record the volume of A used.(bi) concentration of B in mol.dm cube (bii) concentration of A in mol.dm cube (biii) Number of hydrogen in 1.0dm cube of A
(H=1,o=16,k=39, Avogadro constant =6.0*10 power of 23)
Oh, it sounds like we have a precipitation party going on here! Let me break it down for you with a sprinkle of humor:
1. Let's start with the precipitate formed when potassium chloride is added. We know Fe2+ and Hg2 2+ are present, so it's likely one of them is responsible for the party. Here's the net ionic equation:
Fe2+ + 2Cl- -> FeCl2
2. Moving on to the remaining solution after filtering off the first precipitate. When potassium sulfate is added, no precipitate forms. It seems like everyone is getting along and no one wants to crash the party. So there's no need for a net ionic equation here, we'll just let them have their peaceful solution.
3. Finally, when potassium carbonate is added to the remaining solution, a new precipitate forms. It's time for another guest to join the party. Here's the net ionic equation you've been waiting for:
Hg2 2+ + CO32- -> HgCO3 (s)
And there you have it, the net ionic equations for each precipitate formation in this ionic solution extravaganza. Time to put on your lab coat and get ready for more chemical entertainment!
To write the net ionic equations for the formation of each of the precipitates observed, we need to determine the specific reactions that occur.
1. Formation of the precipitate when potassium chloride is added:
Based on the information provided, the ions present in the original solution are Hg2 2+, Ba2+, Fe2+, and another unidentified ion (let's call it X+). When potassium chloride (KCl) is added, a precipitate forms. This indicates that a reaction occurs between X+ and Cl- ions to form an insoluble compound.
The net ionic equation can be written as follows:
X+ (aq) + 2Cl- (aq) → XCl2 (s)
2. No formation of a precipitate when potassium sulfate is added:
After filtering off the precipitate formed above, potassium sulfate (K2SO4) is added to the remaining solution, and no precipitate forms. This suggests that none of the ions present in the original solution react with the sulfate ion (SO4 2-).
No net ionic equation is required here as no reaction occurs.
3. Formation of a precipitate when potassium carbonate is added:
Finally, potassium carbonate (K2CO3) is added to the remaining solution, and a precipitate forms. This indicates a reaction between the remaining ions in the solution (Fe2+ and Ba2+) with carbonate ions (CO3 2-).
The net ionic equations for these reactions are:
Fe2+ (aq) + CO3 2- (aq) → FeCO3 (s)
Ba2+ (aq) + CO3 2- (aq) → BaCO3 (s)
In summary:
1. X+ (aq) + 2Cl- (aq) → XCl2 (s)
2. No reaction
3. Fe2+ (aq) + CO3 2- (aq) → FeCO3 (s)
Ba2+ (aq) + CO3 2- (aq) → BaCO3 (s)
Hg2^+2(aq) + 2Cl^-(aq) ==> Hg2Cl2(s)
iron carbonate is done the same way.