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FrozenT [24]
3 years ago
13

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

Round your answer to 1 decimal place.
Chemistry
1 answer:
madam [21]3 years ago
3 0

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

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In the laboratory you dissolve 19.4 g of potassium acetate in a volumetric flask and add water to a total volume of 125 mL. What
aleksandrvk [35]

The molarity of the potassium acetate solution given the data is 1.584 M

<h3>What is molarity? </h3>

This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of CH₃COOK</h3>
  • Mass of CH₃COOK = 19.4 g
  • Molar mass of CH₃COOK = 98 g/mol
  • Mole of CH₃COOK =?

Mole = mass / molar mass

Mole of CH₃COOK = 19.4 / 98

Mole of CH₃COOK = 0.198 mole

<h3>How to determine the molarity of CH₃COOK</h3>
  • Mole of CH₃COOK = 0.198 mole
  • Volume = 125 mL = 125 / 1000 = 0.125 L
  • Molarity of CH₃COOK = ?

Molarity = mole / Volume

Molarity of CH₃COOK = 0.198 / 0.125

Molarity of CH₃COOK = 1.584 M

Learn more about molarity:

brainly.com/question/15370276

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2 years ago
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Answer:

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The answer would be B.

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The beakers in the picture above are labeled according to the ionic solutions they contain. What metal or metals could replace c
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In such a reaction, metal higher up in the activity series replaces another one due to their position.

To known the metal or metals that will replace the given copper, we need to reference the activity series of metals.

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