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lesantik [10]
2 years ago
11

Calculate the hydronium ion concentration in an aqueous solution with a poh of 4.33 at 25°c.

Chemistry
1 answer:
Jlenok [28]2 years ago
4 0

Taking into account the definition of pH and pOH, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or  hydronium ion H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of hydronium ions</h3>

Being pOH= 4.33, pH is calculated as:

pH + 4.33= 14

pH= 14 - 4.33

<u><em>pH= 9.67</em></u>

Replacing in the definition of pH the concentration of hydronium ions is obtained:

- log [H₃O⁺]= 9.67

Solving:

[H₃O⁺]= 10⁻⁹ ⁶⁷

<u><em>[H₃O⁺]= 2.138×10⁻¹⁰ M</em></u>

Finally, the hydronium ion concentration in an aqueous solution with a pOH of 4.33 at 25°c is 2.138×10⁻¹⁰ M.

Learn more about pH and pOH:

brainly.com/question/16032912

brainly.com/question/13557815

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Goshia [24]

when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm

<h3>Conversion scale</h3>

14.6959 lb/in² = 1 atm

<h3>Data obtained from the question</h3>
  • Pressure (in lb/in²) = 32.5 lb/in²
  • Pressure (in ATM) =?

<h3>How to convert 32.5 lb/in² to atm</h3>

14.6959 lb/in² = 1 atm

Therefore

32.5 lb/in² = 32.5 / 14.6959

32.5 lb/in² = 2.21 atm

Thus, 32.5 lb/in² is equivalent to 2.21 atm

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4 0
2 years ago
What is the percent composition by mass of sulfur in the compound mgso4
Margaret [11]

Answer:

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

Explanation:

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Number of sulfur atom = 1

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100

Sulfur :

=\frac{1\times 32 g/mol}{120 g/mol}\times 100=26.7\%

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

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The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

Now put all the given values in this expression, we get:

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

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