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Calculation of poh

What are the pOH and the pH of a 0.0125- M solution of potassium hydroxide, KOH?

Solution

Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH ] = 0.0125 M :

pOH = −log [ OH ] = −log 0.0125
= ( 1.903 ) = 1.903

The pH can be found from the pOH:

pH + pOH = 14.00
pH = 14.00 pOH = 14.00 1.903 = 12.10

Check your learning

The hydronium ion concentration of vinegar is approximately 4 × 10 −3 M . What are the corresponding values of pOH and pH?

Answer:

pOH = 11.6, pH = 2.4

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The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter ( [link] ).

This figure contains two images. The first, image a, is of an analytical digital p H meter on a laboratory counter. The second, image b, is of a portable hand held digital p H meter.
(a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)

The pH of a solution may also be visually estimated using colored indicators ( [link] ).

This figure contains two images. The first shows a variety of colors of solutions in labeled beakers. A red solution in a beaker is labeled “0.10 M H C l.” An orange solution is labeled “0.10 M C H subscript 3 C O O H.” A yellow-orange solution is labeled “0.1 M N H subscript 4 C l.” A yellow solution is labeled “deionized water.” A second solution beaker is labeled “0.10 M K C l.” A green solution is labeled “0.10 M aniline.” A blue solution is labeled “0.10 M N H subscript 4 C l (a q).” A final beaker containing a dark blue solution is labeled “0.10 M N a O H.” Image b shows pHydrion paper that is used for measuring pH in the range of p H from 1 to 12. The color scale for identifying p H based on color is shown along with several of the test strips used to evaluate p H.
(a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1- M solutions of progressively weaker acids: HCl (pH = l), CH 3 CO 2 H (pH = 3), and NH 4 Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1- M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C 6 H 5 NH 2 (pH = 9), NH 3 (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa)

Key concepts and summary

The concentration of hydronium ion in a solution of an acid in water is greater than 1.0 × 10 −7 M at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than 1.0 × 10 −7 M at 25 °C. The concentration of H 3 O + in a solution can be expressed as the pH of the solution; pH = −log H 3 O + . The concentration of OH can be expressed as the pOH of the solution: pOH = −log[OH ]. In pure water, pH = 7.00 and pOH = 7.00

Key equations

  • pH = −log [ H 3 O + ]
  • pOH = −log[OH ]
  • [H 3 O + ] = 10 −pH
  • [OH ] = 10 −pOH
  • pH + pOH = p K w = 14.00 at 25 °C

Chemistry end of chapter exercises

Explain why a sample of pure water at 40 °C is neutral even though [H 3 O + ] = 1.7 × 10 −7 M . K w is 2.9 × 10 −14 at 40 °C.

In a neutral solution [H 3 O + ] = [OH ]. At 40 °C,
[H 3 O + ] = [OH ] = (2.910 −14 ) 1/2 = 1.7 × 10 −7 .

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The ionization constant for water ( K w ) is 2.9 × 10 −14 at 40 °C. Calculate [H 3 O + ], [OH ], pH, and pOH for pure water at 40 °C.

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The ionization constant for water ( K w ) is 9.311 × 10 −14 at 60 °C. Calculate [H 3 O + ], [OH ], pH, and pOH for pure water at 60 °C.

x = 3.051 × 10 −7 M = [H 3 O + ] = [OH ]
pH = −log3.051 × 10 −7 = −(−6.5156) = 6.5156
pOH = pH = 6.5156

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Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

(a) 0.200 M HCl

(b) 0.0143 M NaOH

(c) 3.0 M HNO 3

(d) 0.0031 M Ca(OH) 2

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Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

(a) 0.000259 M HClO 4

(b) 0.21 M NaOH

(c) 0.000071 M Ba(OH) 2

(d) 2.5 M KOH

(a) pH = 3.587; pOH = 10.413; (b) pH = 0.68; pOH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pH = −0.40; pOH = 14.4

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What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?

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What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?

[H 3 O + ] = 3.0 × 10 −7 M ; [OH ] = 3.3 × 10 −8 M

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Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See [link] for useful information.

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Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See [link] for useful information.

[H 3 O + ] = 1 × 10 −2 M ; [OH ] = 1 × 10 −12 M

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The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10 −6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?

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The hydroxide ion concentration in household ammonia is 3.2 × 10 −3 M at 25 °C. What is the concentration of hydronium ions in the solution?

[OH ] = 3.1 × 10 −12 M

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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