<span>
</span>

<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =

Answer:
The value of the equilibrium constant for reaction asked is
.
Explanation:


![K_{goal}=\frac{[C][O_2]}{[CO_2]}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D%5Cfrac%7B%5BC%5D%5BO_2%5D%7D%7B%5BCO_2%5D%7D)
..[1]
![K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCH_3COOH%5D%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E2%7D)
..[2]
![K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2%5D%5E2%5BO_2%5D%7D)
..[3]
![K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_3%3D%5Cfrac%7B%5BC%5D%5E2%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BCH_3COOH%5D%7D)
[1] + [2] + [3]

( on adding the equilibrium constant will get multiplied with each other)



![K=\frac{[C]^2[O_2]^2}{[CO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC%5D%5E2%5BO_2%5D%5E2%7D%7B%5BCO_2%5D%5E2%7D)
On comparing the K and
:


The value of the equilibrium constant for reaction asked is
.
Pure water would be classified as non-conductor since it does not allow the flow of current into the solution. It does not contains any ions or free electrons so it cannot allow the flow of current into the system. It can allow current only when any impurities is present like salts.
Endpoint . this is when the color of the indicator changes while equivalence is when added titrant is chemically equivalent completely to the analyte sample