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storchak [24]
3 years ago
7

When dissolved in water, koh behaves as an acid that forms k+ and oh- ions. a base that forms ko- and h+ ions. a base that forms

k+ and oh- ions. an acid that forms ko- and h+ ions?
Chemistry
1 answer:
Serggg [28]3 years ago
4 0

Answer:

A base that forms K⁺ and OH⁻ ions.

Explanation:

The KOH is an Arrhenius base.

A is <em>wrong</em>. A base does not form H⁺ ions.

B is <em>wrong</em>. A metal hydroxide forms K⁺ ions, not KO⁻ ions.

D is <em>wrong</em>. The metal forms K⁺ ions, KO⁻ ions.

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Watch the animation and observe the titration process of a standard 0.100 M sodium hydroxide solution with 50.0 \rm ml of a 0.10
yuradex [85]

Answer:

Answers are in the explanation

Explanation:

a. The statments are:

The titration process is based on a chemical reaction.  <em>TRUE. </em>A titration is the chemical reaction between an acid and a base.

The pH of the solution at the equivalence point is 7.  <em>TRUE. </em>In a titration of strong acid with strong base the equivalence point is 7

At the endpoint, the pH of solution is 10 and the color of solution is pink.  <em>FALSE</em>. The endpoint in a strong acid - strong base titrationmust be near to equivalence point at pH = 7.

At the beginning of the titration process, the pH of the solution increases rapidly.  <em>FALSE. </em>At beginning of a titration, the pH of the solution increases slowly.

The chemical reaction involved in an acid-base titration is a neutralization reaction.  <em>TRUE. </em>In the reaction, you are neutralizing an acid (HCl) with a base (NaOH)

Before any base is added to the solution, the pH of the solution is high. <em>FALSE. </em>The addition of a base increases pH that is, in the beginning, low.

In a titration process, the endpoint is reached before the equivalence point.  <em>FALSE. </em>In a titration process, the endpoint is reached in the equivalence point

The pH of the solution changes very slowly at the equivalence point. <em>FALSE. </em>At the equivalence point, the pH changes rapidly.

b. For the reaction

NaOH + HCl → H₂O + NaCl

As the reaction is 1:1 and molarity of both solutions are the same, additions < 100mL of NaOH will stay before the equivalence point, equivalence point will be in 100mL and additions > 100mL of NaOH will stay after the equivalence point.

The conditions are:

10.0mL of 1.00 M NaOH before the equivalence point

150 mL of 1.00 M NaOH after the equivalence point

5.00 mL of 1.00 M NaOH before the equivalence point

50.0 mL of 1.00 M NaOH  before the equivalence point

200 mL of 1.00 M NaOH after the equivalence point

I hope it helps!

7 0
3 years ago
Read 2 more answers
Calculate the molarity of a solution of barium hydroxide if 18.15 ml of it is required for the titration of a 20.00 ml sample of
inysia [295]
The balanced equation for the reaction between Ba(OH)₂ and HCl is as follows;
Ba(OH)₂ + 2HCl ---> BaCl₂ + 2H₂O
stoichiometry of Ba(OH)₂ to HCl is 1:2
the number of HCl moles that have reacted - 0.2452 mol/L x (20.00 x 10⁻³ L)
number of HCl moles reacted = 0.004904 mol
2 mol of HCl reacts with 1 mol of Ba(OH)₂
therefore 0.004904 mol of HCl reacts with - 1/2 x 0.004904 mol of Ba(OH)₂
number of Ba(OH)₂ moles in 18.15 mL - 0.002452 mol
Therefore number of Ba(OH)₂ moles in 1000 mL- 0.002452 mol /(18.15 x 10⁻³ L)
molarity of Ba(OH)₂ is = 0.1351 M
6 0
3 years ago
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