If I add water to oil what will happen
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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-]}](https://tex.z-dn.net/?f=%20K_%7BH%7D%20%3D%20%5Cfrac%7B%20K_%7BW%7D%20%7D%7B%20K_%7BA%7D%20%7D%20%3D%20%5Cfrac%7B%5BHCN%5D%5BOH-%5D%7D%7B%5BCN-%5D%7D%20)
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]}](https://tex.z-dn.net/?f=K_%7BH%7D%20%3D%20%5Cfrac%7B%201x10%5E-14%7D%7B%206.2x10%5E-10%20%7D%20%3D%20%5Cfrac%7B%5BX%5D%5BX%5D%7D%7B%5B0.2-X%5D%7D)
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
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:
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,
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