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Varvara68 [4.7K]
3 years ago
6

What is the new solution concentration when 150. mL of water is added to 200. mL of a 3.55 M HBr solutions.

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

The new concentration is 2.03M

Explanation:

Step 1: Data given

A 200 mL 3.55 M HBr is diluted with 150 mL

Step 2: The dilution

In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume

In this case, the volume of the stock solution is 200 mL

Adding  150 mL  of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL

The dilution factor wll be 350/200 = 1.75

This makes the diluted concentration:

3.55/1.75 = 2.03M

The new concentration is 2.03M

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The density of ethanol, a colorless liquid that is commonly known as grain alcohol, is 0.798 g/ml. calculate the mass of 16.9 ml
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3 years ago
Now, Johnny the pool guy must add enough Na2CO3 to get the pH back up to 7.60. How many grams of Na2CO3 does Johnny need to add
vaieri [72.5K]

Answer:

m = 3.4126 g

Explanation:

First, the question is incomplete but I already put in the comments the rest of the question.

Let's solve the first two questions, and then the actual question you are asking here to give you a better explanation of how to do it.

1) We need the volume of the pool, in this case is easy. Assuming the pool is rectangular, we use the volume of a parallelepiped which is the following:

V = h * d * w

We have the data, but first we will convert the feet to centimeter. This is because is easier to work the volume in cm³ than in feet.

So the height, width and depth of the pool in centimeter are:

h = 32 * 30.48 = 975.36 cm

w = 18 * 30.48 = 548.64 cm

d = 5.3 * 30.48 = 161.54 cm

Now the volume:

V = 975.36 * 548.64 * 161.54

V = 86,443,528.79 cm³ or 86,443,528.79 mL or 86,443.5 L

2) If the pool has a pH of 6.4, the concentration of H+ can be calculated with the following expression:

[H+] = antlog(-pH) or 10^(-pH)

Replacing we have:

[H+] = 10^(-6.4)

[H+] = 3.98x10^-7 M

3) Finally the question you are asking for.

According to the reaction:

Na2CO3 + H+ → 2Na+ + HCO3−

We can see that there is ratio of 1:1 between the H+ and the Na2CO3, so, if we have initially a concentration of 3.98x10^-7 M, the difference between the new concentration of H+ and the innitial, will give the concentration to be added to the pool to raise the pH. Then, with the molecular weight of Na2CO3 (105.98 g/mol) we can know the mass needed.

The new concentration of [H+] is:

[H+] = 10^(-7.6) = 2.58x10^-8 M

The difference of both [H+] will give concentration of Na2CO3 used:

3.98x10^-7 - 2.58x10^-8 = 3.73x10^-7 M

The moles:

moles = 3.73x10^-7 * 86,443.5 = 0.0322 moles

Finally the mass:

m = 0.0322 * 105.98

m = 3.4126 g    

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