pH is offered to define acidity or basicity that substances. Its variety varies indigenous 0 come 14. That is characterized as negative logarithmof hydrogen ion concentration.

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The expression for pH is discussed below.

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" /> …… (1)

Where

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" /> is the concentration the hydrogen ion.

Dissociation reaction the KCN is together follows:

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Cyanide ions thus developed can react v water to type HCN and

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as follows:

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The relation between , and also is expressed by adhering to relation:

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…… (2)

Where,

is the ionic product consistent of water.

is the dissociation constant of base.

is the dissociation constant of acid.

The value of is

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.

The worth of is

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.

Substitute these values in equation (2).

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Solve because that ,

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The expression because that the HCN is together follows:

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left< extO extH^ - ight>}}left< extC extN^ - ight>" /> …… (3)

Consider x to be adjust in equilibrium concentration. Therefore, equilibrium concentrationof

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, HCN and becomes (0.2 – x), x and also x respectively.

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Solving for x,

*

Therefore concentration of hydroxide ion is 0.002 M.

The expression to calculate pOH is together follows:

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" /> …… (4)

Substitute 0.002 M for

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" /> in equation (4).

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The relation between pH and also pOH is together follows:

pH + pOH = 14 …… (5)

Substitute 2.69 because that pOH in equation (4).

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*

Solving for pH,

pH = 11.31

Learn more:

Write the chemical equation responsible for pH of buffer containing and : factor for the acidic and an easy nature that amino acid.

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acids, base and salts

Keywords: pH, pOH, 11.31, 2.69, 14, 0.002 M, Kb, Kw, Ka, 10^-14, 2*10^-5.



CN^-1 is the conjugate basic of the weak acid HCN ( Ka 3 X 10^-9 )so CN^-1 is a solid base and also will hydrolyze, through the reaction:CN^-1 + H2O --> HCN + OH^-1 Theamount is based on the hydrolysis consistent which is Kw / KaHCN = 1 X 10^-14 / 3 X 10^-9 = 3.333 X 10 ^-6 0.2 M CN^-1 will certainly ionize to a small extent ( A ) to develop A quantities of HCN and A quantities of OH^-1 soo Kh = 3.333 X 10^-6 = / 0.2 -A but A A^2 = 0.2 x 3.333 X 10^-6 A^2= 6.67 X 10^-7 A= 8.17 X 10^-4 pOH = -log ( 8.17 X 10^-4 ) = 3.09then ph ispH = 14 - pOHso pH = 14 - 3.09pH = 10.91

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}" />where K,H is the equilibrium consistent of hydrolysis and Kw is equilibrium continuous for water solvation i beg your pardon is equal to 1×10^-14. Therefore,
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}<0.2-X>" />x = 0.001788 mSince the value of OH- is also x, climate OH-=0.001788 m. Consequently,pOH = -log(0.001788) = 2.75pH = 14 - pOH = 14 - 2.75pH = 11.25