Question

This practice question stumped me. What does it mean by calculating delta H for the "base", and how do you solve the question?

When 510mL of a 0.45M Sr(OH)₂ solution is mixed with 730mL of a 0.6M H₃PO₄ solution, the resulting 280g iron calorimeter had an associated temperature increase of 7.5 degrees C. Calculate change in enthalpy (delta H) of the reaction for the base. Assume the solution has a specific heat of 4.18J/gC.

The reaction is as follows: 3Sr(OH)₂ + 2H₃PO₄ --> Sr₃(PO₄)₂ + 6H₂O

Answer 1

V = 510 ml

M = 0.45 Sr(OH)2

V2 = 730 ml

M = 0.6M H3PO4

m = 280g of Iron

dT = 7.5ªC

a) Hrxn for the base

Cp = 4.18

Well...

this is a set of echange of heat

Qwin = the iron calorimeter

Qlost = the solution

Qwin = -Qlost

Qwin = m*Cp*dT (all these data is for Iron)

Qwin = 280*0.450*(7.5) = 945 J were transferred

assume no lost to the media

Therefore

Q reaction = Hrxn = 945

But this must be stated per unit mol of base

3Sr(OH)2 + 2H3PO4 --> Sr3(PO4)2 + 6H2O

calculate moles of base that actually reacted

mol base = M*V = 0.45*0.51 = 0.2295 mol of Base

mol of acid = M*V = 0.6*0.73 = 0.438 mol of acid

there is an excess of acid

3 mol of base are needed to neutralize 2 mol of acid

therefore, we are sure that the base is limiting reactant (and probably why they asked you to do this as a basis)

therefore

Hrxn = 945J per 0.2295 mol o Base

Hrxn = 945/0.2295 = 4117.6 J/mol base