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how find molar enthalpy of neutralization of formic acid

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My teacher gave us an equation: nhcoohdeltaHhcooh+oh + nh30deltaHh30+oh = -(CcaldeltaTcal + cwaterVaciddeltaTacid + cwaterVbasedeltaTbase) & told us to find enthalpy of neutralization fo formic acid.

 

given:

hcooh + oh----h2o + hcoo

h30+oh----2 h2o

 

hcooh: t1=21.4, t2= 27.5, delta T=6.1 (* this was also the delta Tcal)

naoh: 20.4, t2= 27.5 , delta T= 7.1

 

 

 

This is the information I calculated: 1. initial moles - nh30= 0.0632 (0.075mol - 0.011748mol) (which my teacher said to use the equation above with. We had to interpolate deltaHhcooh+oh ( which = -58517)

 

However, I plugged in the values and got -57474.58. Please help. (* how to do the proper procedure)

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my teacher said that the delta T value for cal = delta T acid (which was 6.1)


I also calculated Ccal in a previous question and it was 30.269 which was the value I substituted in this equation above.


I've attached my calc. *Note: the value of 0.063moles is the moles remaining of HCOOH

post-101955-0-76837800-1384047737_thumb.jpg

post-101955-0-54167400-1384047772_thumb.jpg

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