Metal is a conductor of heat
a) before addition of any KOH :
when we use the Ka equation & Ka = 4 x 10^-8 :
Ka = [H+]^2 / [ HCIO]
by substitution:
4 x 10^-8 = [H+]^2 / 0.21
[H+]^2 = (4 x 10^-8) * 0.21
= 8.4 x 10^-9
[H+] = √(8.4 x 10^-9)
= 9.2 x 10^-5 M
when PH = -㏒[H+]
PH = -㏒(9.2 x 10^-5)
= 4
b)After addition of 25 mL of KOH: this produces a buffer solution
So, we will use Henderson-Hasselbalch equation to get PH:
PH = Pka +㏒[Salt]/[acid]
first, we have to get moles of HCIO= molarity * volume
=0.21M * 0.05L
= 0.0105 moles
then, moles of KOH = molarity * volume
= 0.21 * 0.025
=0.00525 moles
∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525
and when the total volume is = 0.05 L + 0.025 L = 0.075 L
So the molarity of HCIO = moles HCIO remaining / total volume
= 0.00525 / 0.075
=0.07 M
and molarity of KCIO = moles KCIO / total volume
= 0.00525 / 0.075
= 0.07 M
and when Ka = 4 x 10^-8
∴Pka =-㏒Ka
= -㏒(4 x 10^-8)
= 7.4
by substitution in H-H equation:
PH = 7.4 + ㏒(0.07/0.07)
∴PH = 7.4
c) after addition of 35 mL of KOH:
we will use the H-H equation again as we have a buffer solution:
PH = Pka + ㏒[salt/acid]
first, we have to get moles HCIO = molarity * volume
= 0.21 M * 0.05L
= 0.0105 moles
then moles KOH = molarity * volume
= 0.22 M* 0.035 L
=0.0077 moles
∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5
when the total volume = 0.05L + 0.035L = 0.085 L
∴ the molarity of HCIO = moles HCIO remaining / total volume
= 8 x 10^-5 / 0.085
= 9.4 x 10^-4 M
and the molarity of KCIO = moles KCIO / total volume
= 0.0077M / 0.085L
= 0.09 M
by substitution:
PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)
∴PH = 8.38
D)After addition of 50 mL:
from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.
the molarity of KCIO = moles KCIO / total volume
= 0.0105mol / 0.1 L
= 0.105 M
when Ka = KW / Kb
∴Kb = 1 x 10^-14 / 4 x 10^-8
= 2.5 x 10^-7
by using Kb expression:
Kb = [CIO-] [OH-] / [KCIO]
when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]
Kb = [OH-]^2 / [KCIO]
2.5 x 10^-7 = [OH-]^2 /0.105
∴[OH-] = 0.00016 M
POH = -㏒[OH-]
∴POH = -㏒0.00016
= 3.8
∴PH = 14- POH
=14 - 3.8
PH = 10.2
e) after addition 60 mL of KOH:
when KOH neutralized all the HCIO so, to get the molarity of KOH solution
M1*V1= M2*V2
when M1 is the molarity of KOH solution
V1 is the total volume = 0.05 + 0.06 = 0.11 L
M2 = 0.21 M
V2 is the excess volume added of KOH = 0.01L
so by substitution:
M1 * 0.11L = 0.21*0.01L
∴M1 =0.02 M
∴[KOH] = [OH-] = 0.02 M
∴POH = -㏒[OH-]
= -㏒0.02
= 1.7
∴PH = 14- POH
= 14- 1.7
= 12.3
When KCL, which is an ionic compound is added to water it will dissociate and or ionize completely forming the ions of K+ and Cl-. The resulting solution would be a neutral solution, as the K+ is a cation of a strong base and Cl- is an anion of a strong acid, and whenever a strong base reacts with a strong acid, a neutral salt is produced
Answer:
<h3>1)</h3>
Structure One:
Structure Two:
Structure Three:
Structure Number Two would likely be the most stable structure.
<h3>2)</h3>
- All five C atoms: 0
- All six H atoms to C: 0
- N atom: +1.
The N atom is the one that is "likely" to be attracted to an anion. See explanation.
Explanation:
When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:
.
For example, for the N atom in structure one of the first question,
- N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
- This N atom is connected to only 1 chemical bond.
- There are three pairs, or 6 electrons that aren't in a chemical bond.
The formal charge of this N atom will be 
.
Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.
<h3>1)</h3>
Structure One:
Structure Two:
Structure Three:
In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.
Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.
<h3>2)</h3>
Similarly,
- All five C atoms: 0
- All six H atoms to C: 0
- N atom: +1.
Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.
Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.
