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katrin [286]
3 years ago
9

If the hydroxide ion concentration of an aqueous solution is 3.0 x 10-4, what is the [H+] in the solution? Is the solution acidi

c, basic, or neutral?
Chemistry
1 answer:
Volgvan3 years ago
6 0

Answer:

[H+] is  3.31 × 10^-11 M

Solution: Basic

Explanation:

In the question we are given;

Concentration of Hydroxide ion [OH-] as 3.0 × 10^-4 M

We are required to determine the [H+] and whether the solution is acidic, basic or neutral.

We know that;

pOH + pH = 14

and that;

pOH = - log [OH-]

Therefore;

pOH = - log ( 3.0 × 10^-4)

       = 3.52

But;

pH = 14 - pOH

Thus;

pH = 14 - 3.52

     = 10.48

But, we know that;

pH = - log[H+]

Therefore;

H+ = Antilog -pH

    = Antilog -10.48

    = 3.31 × 10^-11

Thus, the concentration of H+ ions [H+] is  3.31 × 10^-11 M

Also since the pH is 10.48, then the solution is basic

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___AsCl3+____H2S-->___As2S3+___HCI​
Alex17521 [72]

Answer:

2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl

Explanation:

When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.

We usually balance O and H atoms last.

AsCl₃ + H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 1

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S --- 1

<u>products</u>

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H --- 1

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2 AsCl₃ + H₂S → As₂S₃ +HCl

<u>reactants</u>

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H --- 2

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<u>products</u>

As --- 2

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The number of As atoms is now balanced.

2 AsCl₃ + 3 H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 2

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H --- 6

S --- 3

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

The number of S atoms is now equal on both sides.

2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

<u>products</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

The equation is now balanced.

3 0
2 years ago
The initial pressure and volume of a gas are 55.16 kilopascals and 0.500 liter. According to Boyle’s law, if the volume of the g
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Answer: The final pressure is 34.48kPa

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Initial Volume V1 = 0.500L

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P2 = P1V1/V2

P2 = 55.16*0.5/0.8

P2 = 34.48kPa

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