NH4Br FeCl3 KCH3COO Al(NO3)3 CaF2 BaSO3 H2C6H2O5 (aq) is amphiprotic. Write equa

Question

NH4Br FeCl3 KCH3COO Al(NO3)3 CaF2 BaSO3 H2C6H2O5 (aq) is amphiprotic. Write equations for how it can react as 1) an acid with wqater and 2) a base with water. Describe how the common ion effect is an example of LeChatelier's Principle. Why must the pKa of the weak acid be consider when selecting a conjugate acid-base pair to use when making a buffer at a desired pH? List 3 ways the titration curve of a strong acid-strong base is different from the titration curve of a weak acid-strong base. Describe how to find the equivalence point on a titration curve. The Ksp of AgI is 8.3 times 10-17. What is the solubility (in mol/L) of AgI? (See Sample Exercise 17.11) The Ksp of PbCl2 is 1.7 times 10-5. What is the solubility (in mol/L) of PbCl2? The Ksp of Ca3(PO4)2 is 2.0 times 10-29. What is the solubility (in mol/L) of Ca3(PO4)2? Calculate the solubility of CuBr in a solution that is 0.01 in M KBr. Ksp- of CuBr = 5.3 times 10-12 Calculate the solubility of Ag2CrO4 in a solution that is 0.01 M in AgNO3. Ksp of Ag2CrO4 = 1.210-12 Use LeChatelier's Principle to explain why the solubility of a compound containing a basic anion (that is, the anion of a week acid) increases as the solution becomes more acidic.

Explanation / Answer

I have less time so solving them asap :-

19] H2C6H7O5-

acid ---> HC6H7O52-

base ----> H3C6H7O5

21] because a weak acid will not dissociate completely to give H+ ions in the solution so we need to consider the H+ from water also and from the acid too

22]the titration curve of a this changes at low and high pH values this is the amjor difference b/w the two one in a basic region and the other in a acidic region

23]the equivalence poi nt can be found by the change in colour of the solution by obervation and from the graph

pKa1+ pKa2/2 = equivalence point

24]AgI

Ksp = s^2

s^2 = 8.3*10^-17

solve for s

25]pbCl2

4s^3 = 1.7*10^-5

solve for s

26] 2.0*10^-29 = 27s^3 * 4s^2

= 108s^5 = Ksp

solve for s

27] 5.3* 10^-12 = s* (s+0.01)

neglect s in comparison to 0.01

so we get

s* 0.01 = 5.3* 10^-12

28]do same as no.27

friend i had less time so i solved it for you asap and i couldnt use the calculator to find the value if ny doubt feel free to ask ! happy to help you !

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