Question

(2) Find the oxidation numbers of the bold-labelled atoms in the following species:
(a) CrF4
(b) UO3
(c) HReO4
(d) FeO42-
(e) BrO3-
(f) PH4+

(3) Suppose you have a 2.30 M solution of rubidium hydroxide and you are asked to make up 500.0 mL of 5.00 x 10-3M rubidium hydroxide. How much of the stock solution do you need?
(4) How many grams of sodium formate (NaCHO2) do you need to make 250.0mL of 7.50 x 10-3M solution?
(5) Acetic acid reacts with sodium hydroxide in a neutralization reaction:
HC2H3O2 + NaOH  NaC2H3O2 + H2O
An 10.0 mL acetic acid solution of unknown concentration is titrated with 0.0998M NaOH, and 24.7mL of the NaOH solution is needed to reach endpoint. What is the molarity of the acetic acid solution?
(6) A sample of an europium (II) compound is dissolved and titrated with 0.0521M KMnO4 according to the following equation:
MnO4- + 5Eu2+ + 8H+  Mn2+ + 4H2O + 5Eu3+
And 13.5 mL of the KMnO4 is needed to reach endpoint. What is the mass of Eu in that compound?

Answer 1

2.

a.CrF4= +4 b.UO3=+6 c. HReO4= +7 d. FeO4-2= +6 e. BrO3-= +5 f. ph4+= +5

3 molarity of the stock solotion=2.30M[ c1]

volume of the stock solution needed=?[v1]

molarity of the diluted solution =0.005M[c2]

volume of the solution needed to be prepared= 500ml [v2]

vol of the stock soultion=[v1]= c2 *v2 /c1 = 0.005*500 /2.3 = 1.08 ml.

4. molarity of sodium formate= wt of ch3coona *1000ml / mol wt of ch3coona *volume on ml

0.0075M= wt*1000/ 82*250

wt of ch3coona = 0.153 gm.

5. molarity of naoh=0.0998M [M1]

volume of naOH =24.7 ML[V1]

Volume of acetic acid=10 ml[v2

molarity of acetic acid=[m2]

for a neutralisation reaction M1V1=M2V2

0.0998*24.7= 10* M2

M2=0.246 M.