The amount of i−3(aq)i3−(aq) in a solution can be determined by titration with a solution containing a known concentration of s2o2−3(aq)s2o32−(aq) (thiosulfate ion). the determination is based on the net ionic equation 2s2o2−3(aq)+i−3(aq)⟶s4o2−6(aq)+3i−( aq) 2s2o32−(aq)+i3−(aq)⟶s4o62−(aq)+3i−( aq) given that it requires 25.2 ml25.2 ml of 0.470 m na2s2o3(aq)0.470 m na2s2o3(aq) to titrate a 10.0 ml10.0 ml sample of i−3(aq),i3−(aq), calculate the molarity of i−3(aq)i3−(aq) in the solution.

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Explanation:

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol