A 20.0 mL sample of hydrochloric acid was neutralized with 58.4 mL of 0.250 M NaOH, in a titration.
1 answer:
You might be interested in
Answer : The enthalpy of neutralization is, 56.012 kJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH
Thus, the number of neutralized moles = 0.08 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 400 g
= final temperature of water =
= initial temperature of metal =
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -4480.96 J
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -4480.96 J
n = number of moles used in neutralization = 0.08 mole
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.012 kJ/mole