For each of the following scenarios, one or more substances are added to solutio

Question

For each of the following scenarios, one or more substances are added to solution. Without solving the whole problem, write (a) the pertinent chemical reaction or two, (2) key algebraic relationship and (3) the key chemical/algebraic assumption (if necessary). You may need to look up K_a values in the book table to help you determine the type of key species. (a) 50 mL of 0.85 M KOBr (potassium hypobromite) (b) 150 mL of 0.23 M KOBr to which 1.5 mL of 0.5 M HI added (c) 0.20 M ammonia mixed with 0.5 M ammonium (d) 125 mL of 0.2 M ammonia + 0.5 M ammonium to which 1.5 mL of 0.4 M HCl added (e) 125 mL of 0.2 M ammonia + 0.5 M ammonium to which 1.5 mL of 0.4 M RbOH added (f) 60 mL of 0.25 M Arsenic acid (H_3AsO_3) (g) 80 mL of 0.25 M H_2KAsO_3 (h) 70 mL of 0.28 M NaK_2AsO_3 (i) 40 mL of 0.25 M HK_2AsO_3 (j) 40 mL of 0.25 M NaK_2AsO_3 to which 2.5 mL of 0.3 M HBr added

Explanation / Answer

(a) 50 mL of 0.85 M KOBr

KOBr dissociates into ions as below:

Pertinent chemical equations:

We write the second reaction because OBr- is the conjugate base of the weak acid HOBr and will form HOBr in a aqueous solution.

Key algebraic relationship:

[OBr-] = [KOBr] = 0.85 M

HOBr is a weak acid and the Ka value is tabulated in relevant tables. We can find out the Kb for the conjugate base, OBr- using the relation

Kb = Kw/Ka where Kw = 1.0*10-14

We need to work with Kb since OH- is produced; therefore,

(b) 150 mL of 0.23 M KOBr + 1.5 mL of 0.5 M HI

Pertinent chemical equations:

Key algebraic relationship:

Moles of KOBr = (150 mL)*(1 L/1000 mL)*(0.230 mol/L) = 0.0345 mole.

Mole of HI added = (1.5 mL)*(1 L/1000 mL)*(0.50 mol/L) = 0.00075 mole.

Moles of HOBr formed at equilibrium = 0.00075; moles of KOBr retained at equilibrium = (0.0345 – 0.00075) = 0.03375.

Volume of the solution remains constant; hence we can use moles to write the Henderson-Hasslebach equation as

Key chemical/algebraic assumptions:

1) There was no HOBr present initially, i.e, we discount the dissolution of KOBr in water to form HOBr as shown in (a) above. HOBr is formed only by the action of HI on KOBr.

2) The volume of the solution remains constant and hence the Henderson-Hasslebach equation can be approximated in terms of number of moles of OBr- and HOBr.

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