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serious [3.7K]
2 years ago
8

A student did not read the direction to the experiment properly and mixed up where to place the NaOH solution and the vinegar. H

e put the vinegar in the titrator and the measured amount of NaOH in the beaker. He then added a drop of the phenolphthalein to the solution in the beaker. Does the student need to empty out all the solutions and start over again or can he/she go ahead and run the titration? If he/she run the titration using the solutions as given above, what should he/she expect to see happen for the result?
Chemistry
1 answer:
Mariana [72]2 years ago
3 0

Answer:

At end point there will a transition from pink to colorless.

Explanation:

As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.

The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.

So on addition of phenolphthalein in NaOH the solution will become pink in color.

When we start pouring vinegar from titrator neutralization of NaOH will begin.

On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.

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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. NH4l, CoBr3, Na2
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NaI > Na2SO4 > Co Br3

meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.

Explanation:

The freezing point depression is a colligative property.

That means that it depends on the number of solute particles dissolved.

The formula to calculate the freezing point depression of a solution of a non volatile solute is:

ΔTf = i * Kf * m

Where kf is a constant, m is the molality and i is the van't Hoff factor.

Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.

The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.

As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:

NH4 I → NH4(+) + I(-) => 2 ions

Co Br3 → Co(+) + 3 Br(-) => 4 ions

Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.

So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
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Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}

For example, the mass fraction of Fe is simply:

(2) MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5

If you combine the equations (1) and (2) you have:

(3) H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) Q=m*H_{average}*(T_2-T_1)

Remember that:

<em>Q:</em> Heat gained or lost

<em>m</em>: Mass of the sample you want to analize

H_{average} : The value obtained in equation (3)

T_2: Final temperature of the sample

T_1: Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

  • Heat gained in a system have a positive value
  • Heat lost in a system have a negative value
  • In this problem the sample loses 200 J, for this reason Q=-200J
  • The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
  • The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
  • The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) -200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)

It is clear that the unknown in equation (5) is T_2

The next step is to calculate T_2. Don't forget the signs; these are important.

Key concept: <u>Since the system is loosing heat, the final temperature of the system ( T_2) should be lower than the initial temperature ( T_1 )</u>

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