Chemistry, 17.09.2019 17:30 Isactuallycarter99

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

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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

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

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