A quantity of 2.00 × 10^2mL of 0.862 MHCl is mixed with an equal volume of 0.431 MBa(OH)2 in a c

Step 1: Determine the limiting reactant. The reaction between HCl and Ba(OH)2 is a neutralizatio

A quantity of 2.00 × 10^2mL of 0.862 MHCl is mixed with an...

A quantity of $2.00 \times 10^{2} \mathrm{mL}$ of $0.862 \mathrm{M} \mathrm{HCl}$ is mixed with an equal volume of $0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}$ in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the $\mathrm{HCl}$ and $\mathrm{Ba}(\mathrm{OH})_{2}$ solutions is the same at $20.48^{\circ} \mathrm{C}$, For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is $-56.2 \mathrm{kJ} / \mathrm{mol} .$ What is the final temperature of the mixed solution?

Transcript

00:01 In neutralization, acid and base react to form salt and water.

00:07 In this problem, we're given with a neutralization reaction between hydrochloric acid and barium hydroxide to form barium chloride and water according to this balanced chemical equation.

00:24 The quantities of the acid and base are given as follows.

00:27 So for hcl, the volume for complete neutralization is 2 times 10 to the power of 2, and the concentration is 0 .862 molar.

00:40 For barium hydroxide we have the same volume as hcl but the concentration is different which is 0 .431 molar.

00:54 Now it is known that the nlp of neutralization reaction so that the hnute is equal to negative 56 .2 kilojoules per mole.

01:07 If the initial temperature of the solution that are mixed is equal to 20 .48 degrees celsius and we assume that no heat is lost to the surroundings, we want to find the final temperature of the solution.

01:25 So notice here that the delta h nute is negative.

01:28 This indicates that heat is lost by the reaction and if all that heat is absorbed by the solution, then we want to quantify the amount of the temperature final.

01:38 So since as the result of absorption of the heat from the reaction, the temperature should rise.

01:45 So let us first find the amount of heat of the reaction.

01:49 So for that, we're going to involve, so typically this involves multiplying the most of the water formed times the delta h of neutralization.

02:02 So we can solve for the most of water formed from the most of hcl since that's the one that we have the information about the volume and the polarity so let's find first the most of hcl okay so most of hcl well we have to have that in a separate aside solution here so most of hcl is equal to the volume in liters multiplied by the malarity so you have liters if you have two times a power of 2 that's 200 m l and equivalent that's equivalent to point 2 liters multiplied by of 0 .862 this gives us the most of hcl as 0 .1724 and notice from the balance equation for every two moles of hcl that it that reacts we also have two most of h2 o so that means one mole of hcl reacting for form one mole of h2 so they are equimolar and so the most of h2 o is the same as the most of hcl, which is .1724.

03:23 So now, let's solve for the q reaction.

03:27 So q reaction is equal to, you have the most of water, which is 0 .1724, multiplied by the heat entity of neutralization, which is negative 56 .4, giving us negative 9 .7236 kilojews...

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