Which of these is NOT a type of chemical reaction?
2 answers:
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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
Answer:
No. While gold would not react with a silver nitrate solution, nickel would.
Explanation:
Refer to the metal reactivity series.
Reactivity: .
Gold is positioned after silver in the reactivity series, meaning that gold is typically less reactive than silver. Thus, gold would not react with a solution of silver ions to produce silver metal.
However, since nickel is positioned before silver in the reactivity series, it is expected that nickel would react with silver ions in this solution to produce silver metal.
Thus, if the silver nitrate solution comes into contact with the two rings, the nickel ring would likely react with the solution, the gold ring would not.