Question #74362
55.0 mL of 0.200 M H3PO4 is mixed with 55.0 mL of 0.600 M KOH intially at 21.62C Predict the final temperature of the solution if its density is 1.13 g/ml and its specfic heat is 3.78J/(gC) Assume the total volcume is the sume of the individual volumes
h3 po4(aq)+3koh(aq) to 3h2o(l)+k3po4(aq)+173.2
h3 po4(aq)+3koh(aq) to 3h2o(l)+k3po4(aq)+173.2
Expert's answer
H3PO4 solution contains: 55.0 mL x 0.200 M / 1000 mL = 0.011 mol of acid
KOH solution contains: 55.0 mL x 0.600 M / 1000 mL = 0.033 mol af base (3 times more that the acid, the nessesary quantity to react completely)
According to the reaction equilibrium:
1 mole of acid reacts and produses 173.2 kJ
0.011 moles – x
X = 1.9052 kJ = 1905.2 J.
Q = cm(T2 – T1)
The mass of final sln:
55 mL + 55 mL = 110 mL
110 mL x 1.13 = 124.3 g
1905.2 J = 3.78 J/gC x 124.3 g (T2 - 21.62)
T2 = 25.67 C
Answer: 25.67 C.
KOH solution contains: 55.0 mL x 0.600 M / 1000 mL = 0.033 mol af base (3 times more that the acid, the nessesary quantity to react completely)
According to the reaction equilibrium:
1 mole of acid reacts and produses 173.2 kJ
0.011 moles – x
X = 1.9052 kJ = 1905.2 J.
Q = cm(T2 – T1)
The mass of final sln:
55 mL + 55 mL = 110 mL
110 mL x 1.13 = 124.3 g
1905.2 J = 3.78 J/gC x 124.3 g (T2 - 21.62)
T2 = 25.67 C
Answer: 25.67 C.
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#331389 on Nov 2022
#331389 on Nov 2022
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