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.

Respuesta :

Explanation:

The reaction between hydrogen ion and hydroxide ion will be as follows.

        [tex]H^{+} + OH^{-} \rightarrow H_{2}O[/tex]

so, ratio between hydrogen and hydroxide ions is 1 : 1.

Therefore, moles of [tex]H^{+}[/tex] = volume × concentration of [tex]H^{+}[/tex]

                             = [tex]150 mL \times frac{1 L}{1000 mL} \times 0.4 M[/tex]

                              = 0.06 mol

Similarly, moles of [tex]OH^{-}[/tex] = volume × concentration of [tex]OH^{-}[/tex]

                               = [tex]200 mL \times frac{1 L}{1000 mL} \times 0.5 M[/tex]

                              = 0.1 mol

Therefore, ratio of moles of hydrogen and hydroxide ions is as follows.

                                     0.06 : 0.1

                                  = 0.6 : 1

As, hydroxide ions are present in excess so, hydrogen ions are the limiting reagent.

Hence, moles of water formed = moles of [tex]H^{+}[/tex] ions = 0.06 mol.

So,    heat released = moles of [tex]H_{2}O \times \Delta H_{neutralization}[/tex]

                                = 0.062 × 62.0 kJ/mol

                                = 3.72 kJ

                                = 3.72 × 1000 J

                                = 3720 J

Let T is the initial temperature. So,

              Heat released = Heat absorbed by the solution

                                       = [tex]m \times C \times \Delta T[/tex]

            3720 J = [tex]350 g \times 4.184 j/g ^{0}C \times (26.5 - T)[/tex]

                          T = [tex]24.0 ^{o}C[/tex]

Thus, we can conclude that initial temperature is [tex]24.0 ^{o}C[/tex].                          

The change in the enthalpy due to the reaction between the acid and base to form a neutral salt is called the enthalpy of neutralization.  The initial temperature of the neutralization reaction is [tex]\rm 24 ^{\circ} C[/tex].

What is the relation between heat and temperature?

The formula for heat can be given as, [tex]\rm Q = m \times C \times \times \Delta T[/tex]

The reaction can be shown as:

[tex]\rm H^{+} + OH^{-} \rightarrow H_{2}O[/tex]

From the reaction, it can be seen that the ratio of hydrogen and hydroxide ions = 1 : 1.

So, moles of hydroxide ions:

[tex]\begin{aligned}\rm moles &= \rm volume \times \text{concentration of hydrogen ion}\\\\&= 0.15 \times 0.4 \;\rm M\\\\&= 0.06 \;\rm mol\end{aligned}[/tex]

Similarly, moles of hydroxide ion can be calculated as:

[tex]\begin{aligned}\rm moles &=\rm volume \times \text{concentration of hydroxide ion}\\\\&= 0.2 \times 0.5\;\rm M\\\\&= 0.1 \;\rm mol\end{aligned}[/tex]

Hence, the ratio of moles of hydrogen ions and hydroxide ions = 0.6 : 1

Hydrogen ions are the limiting reagent and the hydroxide ions are in excess.

Thus, moles of water formed = 0.06 mol

Heat released will be calculated as:

[tex]\begin{aligned}\rm heat\;\rm released &= \text{ moles of }\rm H_{2} \times \Delta H_{\rm neutralization}\\\\&= 0.062 \times 62.0 \;\rm kJ/mol\\\\&= 3.72 \times 1000 \;\rm J\\\\&= 3720\;\rm J\end{aligned}[/tex]

When T is the initial temperature then, the heat released = heat absorbed by the solution.

[tex]\begin{aligned}\rm Q &=\rm m \times C \times \times \Delta T\\\\3720 &= 350 \times 4.184 \times (26.5 - \rm T)\\\\&= 24 ^{\circ }\rm C\end{aligned}[/tex]

Therefore, [tex]\rm 24 ^{\circ} C[/tex] is the initial temperature.

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