Question

Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places

Answer 1

The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.

KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be

CN- + H2O ⇔ HCN + OH-

The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.

          CN-    +    H2O    ⇔    HCN +    OH-

I      0.2 m             ∞                   0             0
C      -x                 ∞                   +x        +x
-----------------------------------------------------------
E      0.2-x                               +x         +x

The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:

$K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}$

where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,

$K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}$

x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,

pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25

Answer 2

The pH of a 0.20 M solution of KCN is $\boxed{11.31}$.

Further Explanation:

pH is used to describe acidity or basicity of substances. Its range varies from 0 to 14. It is defined as negative logarithmof hydrogen ion concentration.

The expression for pH is mentioned below.

${\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]$                                                                     …… (1)

Where $\left[ {{{\text{H}}^ + }}\right]$ is the concentration of hydrogen ion.

Dissociation reaction of KCN is as follows:

${\text{KCN}} \to {{\text{K}}^ + } + {\text{C}}{{\text{N}}^ - }$  

Cyanide ions thus formed can react with water to form HCN and ${\text{O}}{{\text{H}}^ - }$ as follows:

${\text{C}}{{\text{N}}^ - } + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{HCN}} + {\text{O}}{{\text{H}}^ - }$  

The relation between ${{\text{K}}_{\text{w}}}$, ${{\text{K}}_{\text{b}}}$ and ${{\text{K}}_{\text{a}}}$ is expressed by following relation:

${{\text{K}}_{\text{w}}} = {{\text{K}}_{\text{b}}} \cdot {{\text{K}}_{\text{a}}}$                                                                                 …… (2)

Where,

${{\text{K}}_{\text{w}}}$ is the ionic product constant of water.

${{\text{K}}_{\text{b}}}$ is the dissociation constant of base.

${{\text{K}}_{\text{a}}}$ is the dissociation constant of acid.

The value of ${{\text{K}}_{\text{w}}}$ is ${10^{ - 14}}$.

The value of ${{\text{K}}_{\text{a}}}$ is $4.9 \times {10^{ - 10}}$.

Substitute these values in equation (2).

${10^{ - 14}} = {{\text{K}}_{\text{b}}}\left( {4.9 \times {{10}^{ - 10}}} \right)$  

Solve for ${{\text{K}}_{\text{b}}}$,

${{\text{K}}_{\text{b}}} = 2 \times {10^{ - 5}}$  

The expression for ${{\text{K}}_{\text{b}}}$ of HCN is as follows:

${{\text{K}}_{\text{b}}} = \dfrac{{\left[ {{\text{HCN}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{C}}{{\text{N}}^ - }} \right]}}$                                                                            …… (3)

Consider x to be change in equilibrium concentration. Therefore, equilibrium concentrationof ${\text{C}}{{\text{N}}^ - }$, HCN and   becomes (0.2 – x), x and x respectively.

${\text{2}} \times {\text{1}}{{\text{0}}^{ - 5}} = \dfrac{{{x^2}}}{{\left( {0.2 - x} \right)}}$  

Solving for x,

$x = 0.002$  

Therefore concentration of hydroxide ion is 0.002 M.

The expression to calculate pOH is as follows:

${\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]$                                                                             …… (4)

Substitute 0.002 M for $\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ in equation (4).

$\begin{aligned}{\text{pOH}} &= - \log \left( {0.002{\text{ M}}} \right) \\&= 2.69 \\\end{aligned}$  

The relation between pH and pOH is as follows:

pH + pOH = 14                                                                          …… (5)

Substitute 2.69 for pOH in equation (4).

${\text{pH}} + 2.69 = 14$  

Solving for pH,

pH = 11.31

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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.