What type of bond will form between the following pairs of atoms?
1 answer:
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Explanation:
As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.
As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.
So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.
Therefore, concentration of chlorine ions will be calculated as follows.
Concentration of ions =
= 0.6 mM
Thus, we can conclude that the concentration of chloride ions is 0.6 mM.
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