what is the poh of a solution that has a oh- concentration equal to 1.3 × 10-10 m
pH, pOH, pKa, and pKb
- Calculating pH
- Calculating hydronium ion concentration from pH
- Computing pOH
- Calculating hydroxide ion concentration from pOH
- The relationship between pH and pOH
- Calculating pKa
- Computing Ga from pKa
- Calculating pKb
- Calculating Grandb from pKb
To calculate the pH of an aqueous solution you lot need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is so calculated using the expression:
pH = - log [H3O+].
Example: Detect the pH of a 0.0025 M HCl solution. The HCl is a potent acid and is 100% ionized in h2o. The hydronium ion concentration is 0.0025 M. Thus:
pH = - log (0.0025) = - ( - 2.60) = 2.60
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Calculating the Hydronium Ion Concentration from pH
The hydronium ion concentration can be found from the pH by the opposite of the mathematical functioning employed to observe the pH.
[H3O+] = 10-pH or [H3O+] = antilog (- pH)
Example: What is the hydronium ion concentration in a solution that has a pH of 8.34?
- 8.34 = log [HthreeO+]
[H3O+] = 10-eight.34 = four.57 ten 10-ix M
On a calculator, summate 10-8.34, or "changed" log ( - 8.34).
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Calculating pOH
To calculate the pOH of a solution you demand to know the concentration of the hydroxide ion in moles per liter (molarity). The pOH is then calculated using the expression:
pOH = - log [OH-]
Example: What is the pOH of a solution that has a hydroxide ion concentration of iv.82 x 10-5 G?
pOH = - log [four.82 x 10-5] = - ( - 4.32) = four.32
Peak
Calculating the Hydroxide Ion Concentration from pOH
The hydroxide ion concentration tin be found from the pOH by the reverse mathematical operation employed to detect the pOH.
[OH-] = x-pOH or [OH-] = antilog ( - pOH)
Case: What is the hydroxide ion concentration in a solution that has a pOH of 5.70?
v.70 = - log [OH-]
-5.70 = log[OH-]
[OH-] = 10-5.70 = 2.00 ten x-half dozen 1000
On a computer calculate x-5.lxx, or "inverse" log (- v.70).
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Human relationship Between pH and pOH
The pH and pOH of a water solution at 25oC are related by the post-obit equation.
pH + pOH = 14
If either the pH or the pOH of a solution is known, the other tin can be quickly calculated.
Case: A solution has a pOH of 11.76. What is the pH of this solution?
pH = xiv - pOH = 14 - 11.76 = ii.24
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Calculating pKa
The pKa is calculated using the expression:
pKa = - log (Ka)
where "1000a" is the equilibrium abiding for the ionization of the acrid.
Example: What is the pKa of acerb acid, if Ka for acerb acid is 1.78 10 10-5?
pKa = - log (1.78 x x-5) = - ( - 4.75) = four.75
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Calculating Ka from the pKa
The Ka for an acid is calculated from the pKa by performing the opposite of the mathematical functioning used to find pKa.
Ka = ten-pKa or Chiliada = antilog ( - pKa)
Example: Calculate the value of the ionization constant for the ammonium ion, Ka, if the pKa is 9.74.
9.74 = - log (Thoua)
-9.74 = log (Ka)
Ma = 10-ix.74 = 1.82 ten 10-10
On a calculator summate x-9.74, or "changed" log ( - nine.74).
Pinnacle
Calculating pKb
The pKb is calculated using the expression:
pKb = - log (Thoub)
where Kb is the equilibrium constant for the ionization of a base of operations.
Example: What is the pKb for methyl amine, if the value of Kb for methyl amine is 4.4 x 10-iv?
pKb = - log (4.4 10 10-iv) = - ( - three.36) = 3.36
Acme
Calculating Kb from pKb
The Kb for an acrid is calculated from the pKb by performing the opposite of the mathematical operation used to find pKb.
Kb = 10-pKb or Thoub = antilog ( - pKb)
Example: Calculate the value of the ionization constant, Kb, for aniline if the pKb is 9.38.
9.38 = - log (Kb)
-9.38 = log (Kb)
Kb = 10-ix.38 = 4.17 ten 10-10
On a calculator summate 10-9.38, or "inverse" log ( - ix.38).
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Source: https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm
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