A sample of pure tin metal is dissolved in nitric acid to produce 15.00 mL of solution containing Sn2+. When this tin solution is titrated, a total of 42.1 mL of 0.145 mol/L KMnO4 is required to reach the equivalence point. a. What is the concentration of the Sn2+ solution?b. Find the concentration of the Sn2+(aq) in mol/L: (give your answer to 3 decimal places)

1.00 M

Explanation:

Sn^2+ reacts with KMNO4 as follows;

5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) > 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)

The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L

= 0.0061 moles

If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-

x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-

x= 5 × 0.0061/2

x= 0.015 moles

Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L

number of moles = concentration × volume

Concentration = number of moles/volume

Concentration= 0.015 moles/0.015 L

Concentration = 1 M

neutrons are in the middle of the atom and the electrons, which are much smaller.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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