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natita [175]
3 years ago
6

Calculate the concentration of h3o⁺ in a solution that contains 5.5 × 10-5 m oh⁻ at 25°c. Identify the solution as acidic

Chemistry
1 answer:
Andreas93 [3]3 years ago
7 0

Answer: [H_3O^+]=1.6\times 10^{-10}M , the solution is basic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

pH=-\log [H_3O^+]

pOH=-\log[OH^-]

Putting in the values:

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

pOH=4.2

pH+pOH=14

pH=(14-pOH)=(14-4.2)=9.8

9.8=-\log [H_3O^+]

[H_3O^+]=1.6\times 10^{-10}M

As pH is more than 7, the solution is basic

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Read 2 more answers
What is the osmotic pressure, in torr, of a 3.00% solution of NaCl in water when the temperature of the solution is 45 ºC? Enter
Anit [1.1K]

213034 torr is the osmotic pressure.

Explanation:

osmotic pressure is calculated by the formula:

osmotic pressure= iCrT

where i= no. of solute

c= concentration in mol/litre

R= Universal Gas constant

T = temp

It is given that solution is 3% which is 3gms in 100 ml.

let us calculate the concentration in moles/litre

3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl

= 5.372 gm/litre

Putting the values in the formula,                   Temp in Kelvin 318.5K

osmotic pressure= 2*5.372*0.083 * 318.5                 Gas constant  0.083

                              = 284.023 bar or 213018 torr.                 c=  5.372 moles/L                                                                              

                                                                                    i=2 for NaCl

4 0
3 years ago
If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?
aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

    Mn(OH)₂   →    Mn⁺² + 2OH⁻

    [Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

    Ksp = [Mn⁺²][OH-]²

    [Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

    [OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

    [OH⁻]² = 10⁻³

    [OH⁻] = (10⁻³)¹⁽²

    [OH⁻] = 0.0316 M

Calculate the pOH

    pOH = -log [OH⁻]

    pOH =  -log [0.0316]

    pOH = 1.5

Now calculate pH

   pH = 14 - pOH

   pH = 14 - 1.5

   pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0
1 year ago
Please answer the question
Tom [10]

Answer:

Stamens

Explanation:

7 0
3 years ago
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