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3 years ago

What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?

Chemistry

1 answer:

rewona [7]3 years ago

5 0

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

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What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?

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Answer:

295.7 mL of 24% trichloroacetic acid (tca) is needed .

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3 years ago

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3 years ago

For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

3 0

3 years ago