100.0 ml of a 1.18 M acid, HA, is mixed with 100.0 ml of a 1.18 M base, BOH, in
ID: 1056122 • Letter: 1
Question
100.0 ml of a 1.18 M acid, HA, is mixed with 100.0 ml of a 1.18 M base, BOH, in a calorimeter whose calorimeter constant is 37.1J/ o C, making a 200.0 ml AB salt solution.
HA(aq) + BOH(aq) AB(aq) + H2O(l)
The temperature of the solution and calorimeter increased from 19.7 oC to 27.5 oC. Assume that the density and specific heat of the salt solution is the same as that of water, which is the dominant component.
Note: Water: d = 1.00 g/ml s = 4.184 J/(g o C)
A) How much heat was absorbed by the solution?
B) How much heat was absorbed by the calorimeter?
A) ____________________J
B) ____________________J
Explanation / Answer
A)
volume = 200 mL
density = 1 g/mL
m = 200 g
Q = m*Cw*delta T
= 200*4.184*(27.5 - 19.7)
= 6527 J
Answer: 6527 J
B)
Q = C*delta T
= 37.1**(27.5 - 19.7)
= 289.4 J
Answer: 289.4 J
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