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When 150. Ml of 0. 400 m h+ are mixed with 200. Ml of 0. 500 m oh-, the final temperature of the solution is 26. 5°c. What was the initial temperature of the solution before the reaction occurred? assume that the solution has a total mass of 350. G and a specific heat capacity of 4. 184 j/g°c. The enthalpy of neutralization for the reaction is -62. 0 kj/mol of water produced.

Sagot :

The initial temperature of the solution before the reaction occurred is; 24°C

Stoichiometry Calculations

The reaction between the hydrogen ion and hydroxide ion will be;

H⁺ + OH⁻ → H₂O

We can see that the ratio of hydrogen ions to hydroxide ions is 1:1.

Thus,

No. of moles of H⁺ = volume × concentration of H⁺

We are given;

Volume of H⁺ = 150 mL = 0.15 L

Concentration of H⁺ = 0.4 M

Thus;

No. of moles of H⁺ = 0.15 * 0.4

No. of moles of H⁺ = 0.06 mol

In a similar fashion;

No. of moles of OH⁻ = volume × concentration of OH⁻

We are given;

Volume of OH⁻ = 200 mL = 0.2 L

Concentration of OH⁻ = 0.5 M

Thus;

No. of moles of OH⁻ = 0.2 * 0.5

No. of moles of OH⁻ = 0.1 mol

Thus, the ratio of number of moles of hydrogen ions to number of moles of hydroxide ions is 0.06:0.1

Now, hydroxide ions (OH⁻) are present in excess and as such hydrogen ions (H⁺) will be the limiting reagent.

Thus;

No. of moles of water formed = No. of moles of  ions = 0.06 mol.

Thus;

heat released = moles of H⁺ ions * enthalpy of neutralization(ΔH_neutr)

heat released = 0.06 mol * -62 kj/mol

heat released = 3.72 kJ = 3720 J

From energy balance, we know that;

Amount of heat released = Amount of heat absorbed by the solution

Thus;

3720 = m*c*ΔT

3720 = 350 * 4.184 * (26.5 - T)

where T is the initial temperature.

Solving for T gives T = 24°C

Read more about Stoichiometry at; https://brainly.com/question/16060223

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