Answer:
It's spontaneous in the reverse direction
Explanation:
A negative voltage indicate s that the reverse reaction is spontaneous (i.e. oxidation at the cathode, and reduction at the anode; by convention you would need to swap the labels on the electrodes)
Stir the water continuously, this is the only logical answer. Adding powdered sugar, decreasing the volume, increasing the amount of suga, cooling, all don't make the sugar dissolve quicker.
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
Answer:
B. Valence Electron Pairs
Explanation:
Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.
