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.

" /> …… (1)Where

" /> is the concentration the hydrogen ion.Dissociation reaction the KCN is together follows:

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

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

…… (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

.The worth of is

.Substitute these values in equation (2).

Solve because that ,

The expression because that the HCN is together follows:

left< extO extH^ - ight>}}left< extC extN^ - ight>" /> …… (3)Consider x to be adjust in equilibrium concentration. Therefore, equilibrium concentrationof

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

Therefore concentration of hydroxide ion is 0.002 M.

The expression to calculate pOH is together follows:

" /> …… (4)

Substitute 0.002 M for

" /> in equation (4).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