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irinina [24]
3 years ago
13

Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic

, or neutral.
a. 1.8 × 10-10 M, acidic
b. 9.2 × 10-1 M, basic
c. 5.5 × 10-10 M, neutral
d. 9.2 × 10-1 M, acidic
e. 1.8 × 10-10 M, basic
Chemistry
1 answer:
defon3 years ago
6 0

Answer:

The correct answer is option (e).

Explanation:

Concentration of hydroxide ions [OH^-]= 5.5\times 10^{-5} M

The pOH of the solution is defined as negative logarithm of hydroxide ions of in an aqueous solution.

  • Higher the value of pOH more will be the acidic solution.
  • Lower the value of pOH more will be the basic solution.
  • At value  equal to 7 the solution is said to be neutral.

pOH=-\log[OH^-]

pOH=-\log[5.5\times 10^{-5} M]=4.259

pH +pOH = 14

pH = 14-pOH

pH=14 - 4.259 = 9.741

The pH of the solution is defined as negative logarithm of hydronium ions of in an aqueous solution.

  • Higher the value of pH more will be the basic solution.
  • Lower the value of pH more will be the acidic solution.
  • At value equal to 7 the solution is said to be neutral.

pH=-\log[H_3O^+]

9.741=-\log[H_3O^+]

[H_3O^+]=1.8\times 10^{-10} M

The solution is basic as pH of the solution is 9.741.

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3 0
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Read 2 more answers
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
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