Answer:
an ion
Explanation:
An ion is an atom or molecule with a net electric charge due to the loss or gain of one or more electrons. This is not a particle within an atom
Answer:
a. True
Explanation:
The data curve for OCI-/H3O appears to be similar to C2H3O2-/H3O because protons have not dissociated with conjugate base. This conjugate base is strong and hold proton in solution while acids try to dissociate it.
Answer:
i) increase
ii) decrease
iii) remain the same
iv) No, because it dissociates completely.
Explanation:
On a 10-fold dilution of a weak acid, the pH will increase because the concentration of hydrogen ions will decrease thereby increasing the pH to close to that of water.
On a 10-fold dilution of a weak base, the pH will decrease due to the removal of hydroxide ions from the solution. This results in the solution having a H closer to that of water.
If one adds a very small amount of strong base to a buffered solution, the pH will remain constant because a buffer solution acts to withstand any change to its pH on the addition of small quantities of either an acid or a base.
A buffer solution cannot be made with a strong acid because thy undergo complete dissociation. Therefore, any small addition of base or acid will result in very large changes in the pH of the solution. A buffer solution is made with a weak acid and its conjugate base or a weak base and its conjugate acid.
Answer: The pH at the equivalence point for the titration will be 0.65.
Solution:
Let the concentration of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
be x
Initial concentration of ![[CH3NH_2]](https://tex.z-dn.net/?f=%5BCH3NH_2%5D)
, c = 0.230 M
at eq'm c-x x x
Expression of 
:
![K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5B%2BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7Bc-x%7D%3D%5Cfrac%7Bx%5E2%7D%7Bc-x%7D)
Since ,methyl-amine is a weak base,c>>x so 
.
Solving for x, we get:
Given, HCl with 0.230 M , it dissociates fully in water which means ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
= 0.230 M
![[OH^-]=[H^+]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BH%5E%2B%5D)
will result in neutral solution, since ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3C%5BH%5E%2B%5D)
Remaining after neutralizing ions
![[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D%5BH%5E%2B%5D-%5BOH%5E-%5D%3D0.230-1.07%5Ctimes%2010%5E%7B-2%7D%3D0.2193%20M)
![pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65](https://tex.z-dn.net/?f=pH%3D-log%7B%5BH%5E%2B%5D_%7B%5Ctext%7Bleft%20in%20solution%7D%7D%3D-log%280.2193%29%3D0.65)
The pH at the equivalence point for the titration will be 0.65.
