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Ksivusya [100]
4 years ago
9

Discribe how to prepare 50.00mL of a 3.00M hydrochloric acid solution using 12.0 M HCl.

Chemistry
1 answer:
Leokris [45]4 years ago
6 0

Answer:

The answer to your question is given below.

Explanation:

To prepare 50mL of 3M HCl, we must calculate the volume of the stock solution needed. This can obtained as follow:

Molarity of stock solution (M1) = 12M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 3M

Volume of diluted solution (V2) = 50mL

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12 x V1 = 3 x 50

Divide both side by 12

V1 = (3 x 50)/12

V1 = 12.5mL

The volume of the stock solution needed is 12.5mL

Therefore, to prepare 50mL of 3M HCl, we must measure 12.5mL of the stock solution i.e 12M HCl and then, add water to the mark in a 1L volumetric flask. Now we can measure out 50mL of the solution.

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Please helpWhat’s the pH of a solution of ammonia that has a concentration of 0.335 M? The Kb of ammonia is
natta225 [31]

From the calculation, the pH of the solution is 4.85.

<h3>What is the pH?</h3>

The pH is defined as the hydrogen ion concentration of the solution. We have the ICE table as;

         HA    +    H2O ⇔     H3O^+    +    A^-

I       0.335                          0                0

C     -x                                  +x              +x

E   0.335 - x                         x                x

Ka = 1 * 10^-14/Kb

Ka = 1 * 10^-14/1.8 × 10^–5

Ka = 5.56 * 10^-10

Ka = [H3O^+] [A^-]/[HA]

But  [H3O^+] = [A^-] = x

5.56 * 10^-10 = x^2/ 0.335 - x

5.56 * 10^-10(0.335 - x ) =  x^2

1.86  * 10^-10 - 5.56 * 10^-10x =  x^2

x^2 + 5.56 * 10^-10x - 1.86  * 10^-10 = 0

x=0.000014 M

Now;

pH = -log 0.000014 M

pH = 4.85

Learn more about pH:brainly.com/question/15289741

#SPJ1

7 0
2 years ago
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