alculate the pH of the following solutions. Part A Solution prepared by dissolvi

Question

alculate the pH of the following solutions.

Part A

Solution prepared by dissolving 5.5g of lithium hydroxide in water to give 230mL of solution.

Express your answer using two decimal places.

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Part B

Solution prepared by dissolving 0.96g of hydrogen chloride in water to give 0.48L of solution.

Express your answer using two decimal places.

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Part C

Solution prepared by diluting 57.0mL of 0.120M HCl to a volume of 1.00L .

Express your answer using two decimal places.

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Part D

Solution prepared by mixing 119.5mL of 1.3

alculate the pH of the following solutions.

Part A

Solution prepared by dissolving 5.5g of lithium hydroxide in water to give 230mL of solution.

Express your answer using two decimal places.

pH =

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Part B

Solution prepared by dissolving 0.96g of hydrogen chloride in water to give 0.48L of solution.

Express your answer using two decimal places.

pH =

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Part C

Solution prepared by diluting 57.0mL of 0.120M HCl to a volume of 1.00L .

Express your answer using two decimal places.

pH =

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Part D

Solution prepared by mixing 119.5mL of 1.3

Explanation / Answer

Part A:

5.5 g of LiOH in 230 ml of solution.

Moles of LiOH = Mass / Molar mass = 5.5 / 23.95 = 0.23 mol

Volume = 230 ml = 0.230 L

Now, Molarity = Moles of solute / Volume of solution in L = 0.23 mol / 0.230 L = 1 M

LiOH = 1 M

Li+ = 1M and OH- = 1M

pOH = -log [OH^-]

=> pOH = -log[1]

=>pOH = 0

pH = 14 - pOH

=> pH = 14.00

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PArt B:

0.96 g of HCl in 0.48 L of solution.

Moles of HCl = Mass / Molar mass = 0.96 / 36.5 = 0.026 mol

Volume = 0.48 L

Now, Molarity = Moles of solute / Volume of solution in L = 0.026 mol / 0.48 L = 0.054 M

HCl = 0.054 M

H+ = 0.054 M and Cl- = 0.054M

pH = -log [H^+]

=> pH = -log[0.054]

=>pH = 1.27

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Part C:

Diluting 57.0 ml of 0.120 M HCl to a volume of 1.00L

Moles of HCl in 57.0 ml of 0.120 M HCl = Molarity x Volume (in L) = 6.84 x 10^-3 mol

Final volume of solution = 1.00 L

Molarity of final solution = Molarity of solute / Volume of solution = 6.84 x 10^-3 mol / 1.00 = 6.84 x 10^-3 M

pH = -log [H^+]

=> pH = -log (6.84 x 10^-3 mol)

=>pH = 2.21

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Part D:

119.5 ml of 1.3 x 10^-3 M HCl and 416.5 ml of 1.4 x 10^-3 M HClO4

Moles of HCl = Molarity x Volume (in L) = 1.3 x 10^-3 x 0.1195 = 1.55 x 10^-4 mol

Final volume = 536 mL = 0.536 L

Molarity of HCl = Moles of HCl / Final volume = 1.3 x 10^-3 / 0.536 = 2.43 x 10^-3 M

H+ = 2.43 x 10^-3 M

Moles of HClO4 = Molarity x Volume (in L) = 1.4 x 10^-3 x 0.4165 = 5.83 x 10^-4 M

Molarity of HClO4 = Moles of HClO4 / Final volume = 5.83 x 10^-4 / 0.536 =1.1 x 10^-3 M

H+ = 1.1 x 10^-3 M

Total H+ = 2.43 x 10^-3 M + 1.1 x 10^-3 M = 3.53 x 10^-3 M

pH = -log [H+]

=> pH = -log (3.53 x 10^-3 )

=> pH = 2.45

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