If 50.00mL of 1.10M NaOH is added to 25.00mL of 1.86M HCl, with both solutions originally at 24.70 degrees Celsius, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02g/mL and a specific heat of 3.98J/g*degrees Celcius.)

Express answer in four significant figures.

You have a great deal of information but you're still missing the heat of reaction (heat of neutralization).

oops, here it is:

-55.84 kJ/mol H20 produced

50.00 x 1.10 = 55.00 millimols NaOH

25.00 x 1.86 = 46.5 mmols HCl

.......NaOH + HCl ==> NaCl + H2O
I......55.00..46.50.....0......0
C.....-46.5..-46.5.....46.5...46.5
E......8.5.....0.......46.5...46.5
That 46.5 millimols = 0.0465 mols H2O.
55.84 kJ/mol x 0.0465 mols = 2.597 kJ heat = q.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

I would change q to J from kJ, use density to find mass of the 75.00 mL H2O, substitute Ti and sp.h. and solve for Tf. (I am ignoring the specific heat of the unreacted NaOH(solid) as well as that of the NaCl(solid) formed.)

To solve this problem, we need to use the heat transfer equation:

q = mcΔT

where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.

First, we need to calculate the total mass of the solution. The volume and density information is given, so we can use the formula:

mass = volume x density

For the NaOH solution:
mass_NaOH = 50.00 mL x 1.02 g/mL = 51.00 g

For the HCl solution:
mass_HCl = 25.00 mL x 1.02 g/mL = 25.50 g

Next, we calculate the total heat transferred by adding the heat transferred by each solution:

q_total = q_NaOH + q_HCl

The heat transferred by each solution is given by:

q = mcΔT

For the NaOH solution:

q_NaOH = (51.00 g)(3.98 J/g*°C)(ΔT)

For the HCl solution:

q_HCl = (25.50 g)(3.98 J/g*°C)(ΔT)

Since the two solutions mix and react, the total heat transferred is zero (assume no heat is lost to the surrounding air). Therefore:

0 = q_total = q_NaOH + q_HCl

We can set up an equation to solve for the change in temperature, ΔT:

(q_NaOH + q_HCl ) = [(51.00 g)(3.98 J/g*°C)(ΔT)] + [(25.50 g)(3.98 J/g*°C)(ΔT)]

Now, we can solve for ΔT:

0 = (51.00 g)(3.98 J/g*°C)(ΔT) + (25.50 g)(3.98 J/g*°C)(ΔT)

0 = (204.098 g·°C)(ΔT)

ΔT = 0 / (204.098 g·°C)

ΔT = 0 °C

Finally, we calculate the final solution temperature by adding the initial temperatures of the two solutions and dividing by 2:

T_final = (T_NaOH + T_HCl) / 2

T_final = (24.70 °C + 24.70 °C) / 2

T_final = 24.70 °C

Therefore, the final solution temperature will be 24.70 °C.