The neutralization of H3PO4 with KOH is exothermic.
H3PO4(aq) + 3KOH(aq) -> 3H2O(l) +K3PO4(aq) +173.2KJ
70.0 mL of 0.207 M H3PO4 is mixed with 70.0 mL of 0.620 M KOH initially at 21.87 °C. Predict the final temperature of the solution if its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C) Assume that the total volume is the sum of the individual volumes.
You have 70.0mL x 0.207 M = 14.49 millimols which will use 43.47 mmols KOH.
You have 70.0mL x 0.620M = 43.4 mmols KOH. So the reaction will produce 43.4 mmols (0.0434 mols product) or
173,200 J/mol x 0.0434 = ? J.
?J = mass soln x specific heat soln x (Tfinal-Tinitial)
mass soln = (70mL+70mL)*1.13 g/mL
Solve for Tfinal.
I got 7516.88J for ?J
and mass of soln = 158.2g
So my equation is
7516.88 J = 158.2g x 3.78 J/(g·°C)x(Tfinal -21.87)
12.57 = Tfinal - 21.87
Tfinal = 34.44°C
Which was wrong.
The hint they gave me was
Start by finding the amount of heat produced when 70.0 mL of 0.207 M H3PO4 react with 70.0 mL of 0.620 M KOH. Use the total volume and the density to find the mass. Then use the mass, m, heat, q, and specific heat, c, to find the change in temperature, ΔT.
Any idea as to what went wrong?
Well, it seems like you need a little help with your chemistry problem. Don't worry, I'm here to assist you! But before I dive into the answer, let me give you a little humor to lighten the mood.
Why did the chemist like nitrates so much?
Because they're cheaper than day rates!
Now, let's get down to business. In order to solve this problem, we'll need to use the equation q = m·c·ΔT, where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to calculate the mass of the solution. Since density is given as 1.13 g/mL and the total volume is the sum of the individual volumes, we can find the mass by multiplying the volume by the density. The total volume is 140.0 mL, so the mass is 140.0 g.
Next, we need to determine the heat released in the reaction. The stoichiometric coefficients in the balanced equation tell us that the heat released is 173.2 kJ per mole of H3PO4 reacted. To calculate the moles of H3PO4, we can use the formula Molarity = moles/volume. Plugging in the values, we find that the moles of H3PO4 is 0.207 mol.
Now that we have the moles, we can calculate the heat released using the equation: q = moles × heat per mole. Substituting the values, we get q = 0.207 mol × (-173.2 kJ/mol) = -35.87 kJ.
Finally, we can use the equation q = m·c·ΔT to find the change in temperature. Rearranging the equation, we get ΔT = q/(m·c). Substituting the values, we have ΔT = (-35.87 kJ)/(140.0 g × 3.78 J/(g·°C)) ≈ -0.069 °C.
Since the initial temperature was 21.87 °C, we add the change in temperature to find the final temperature: 21.87 °C - 0.069 °C ≈ 21.80 °C.
So, the final temperature of the solution is approximately 21.80 °C. Keep in mind that this is an estimation, and actual values may differ due to experimental variations. I hope this helps, and if you have any more questions, feel free to ask!
To predict the final temperature of the solution, we need to use the concept of heat transfer and the equation
q = m × C × ΔT
where:
- q is the heat transferred (in Joules),
- m is the mass of the solution (in grams),
- C is the specific heat capacity of the solution (in J/(g·°C)), and
- ΔT is the change in temperature (in °C).
1. Calculate the mass of the solution:
Mass = density × volume
Mass = 1.13 g/mL × (70.0 mL + 70.0 mL)
Mass = 1.13 g/mL × 140.0 mL
Mass = 1.13 g/mL × 140.0 cm³
Mass = 158.2 g
2. Calculate the moles of H3PO4 and KOH:
Moles = concentration × volume
Moles H3PO4 = 0.207 M × 70.0 mL
Moles H3PO4 = 0.207 mol/L × 0.0700 L
Moles H3PO4 = 0.0145 mol
(Similarly, calculate the moles of KOH.)
3. Determine the limiting reactant:
To find the limiting reactant, compare the moles of each reactant with their stoichiometric coefficients.
H3PO4: 0.0145 mol
KOH: (Calculate the moles of KOH in a similar manner.)
Divide the moles of each reactant by their respective stoichiometric coefficients to get moles per coefficient.
H3PO4: 0.0145 mol / 1 = 0.0145 mol
KOH: (Calculate moles per coefficient for KOH.)
The reactant with the smaller moles per coefficient value is the limiting reactant.
4. Calculate the heat released by the reaction:
Use the stoichiometry of the balanced equation to find the moles of the limiting reactant that reacted completely and the corresponding heat released.
Moles H3PO4 reacted = 0.0145 mol × 1 / 1 = 0.0145 mol (since H3PO4 has a coefficient of 1)
Heat released = Moles H3PO4 reacted × ΔH
Heat released = 0.0145 mol × 173.2 kJ/mol
Heat released = 2.509 kJ
5. Calculate the change in temperature:
Rearrange the equation: q = m × C × ΔT → ΔT = q / (m × C)
ΔT = 2.509 kJ / (158.2 g × 3.78 J/(g·°C))
ΔT = 2.509 × 10³ J / (158.2 g × 3.78 J/(g·°C))
ΔT = 4.177 °C
6. Determine the final temperature:
The final temperature is the initial temperature plus the change in temperature.
Final temperature = 21.87 °C + 4.177 °C
Final temperature = 26.047 °C
Therefore, the predicted final temperature of the solution would be approximately 26.047 °C.