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3 years ago

658 mL of 0.250 M HCl solution is mixed with 325 mL of 0.600 M HCl solution. What is the molarity of the resulting solution

1 answer:

3 0

The molarity of the resulting solution obtained by mixing 658 mL of 0.250 M HCl solution with 325 mL of 0.600 M HCl solution is 0.366 M

We'll begin by calculating the number of mole of HCl in each solution. This can be obtained as follow:

<h3>For solution 1:</h3>

Volume = 658 mL = 658 / 1000 = 0.658 L

Molarity = 0.250 M

Mole = Molarity x Volume

Mole of HCl = 0.250 × 0.658

<h3>For solution 2:</h3>

Volume = 325 mL = 325 / 1000 = 0.325 L

Molarity = 0.6 M

Mole = Molarity x Volume

Mole of HCl = 0.6 × 0.325

Next, we shall determine the total mole of HCl in the final solution. This can be obtained as follow:

Mole of HCl in solution 1 = 0.1645 mole

Mole of HCl in solution 2 = 0.195 mole

Total mole = 0.1645 + 0.195

<h3>Total mole = 0.3595 mole</h3>

Next, we shall determine the total volume of the final solution.

Volume of solution 1 = 0.658 L

Volume of solution 2 = 0.325 L

Total Volume = 0.658 + 0.325

<h3>Total Volume = 0.983 L</h3>

Finally, we shall determine the molarity of the resulting solution.

Total mole = 0.3595 mole

Total Volume = 0.983 L

<h3>Molarity =?</h3>

Molarity = mole / Volume

Molarity = 0.3595 / 0.983

<h3>Molarity = 0.366 M</h3>

Therefore, the molarity of the resulting solution is 0.366 M

Learn more: brainly.com/question/25342554

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Draw the conjugate base for the Brønsted-Lowry acid-base reaction that occurs when the following acid reacts with water. Show al

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Answer:

The structures are shown below.

Explanation:

When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.

The formal charge of an atom can be calculated by:

FC = X - (Y + Z/2)

Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.

a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.

b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.

c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.

The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.

The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.

The structure is shown in figure c.