The pH of an acid is lower than the pH of a base. On the pH scale, 7 is neutral, 1-6 is considered acidic, and 8-14 is considered alkaline.
A) according to this reaction:
by using ICE table:
NH2OH(aq) + H2O(l) → HONH3+(aq) + OH-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X
when Kb = [OH-][HONH3+]/[NH2OH]
when we have Kb = 1.1x10^-8 so,
by substitution:
1.1x10^-8 = X^2/(0.4-X) by solving this equation for X
∴X = 6.6x10^-5 M
∴[OH] = 6.6x10^-5 M
when POH = - ㏒[OH]
∴POH = -㏒(6.6x10^-5)= 4.18
∴PH = 14 - POH = 14 - 4.18
= 9.82
when PH = -㏒[H+]
∴[H+] = 10^9.82 = 1.5x10^-10 M+0.02molHcl
= 0.02
∴ the new value of PH = -㏒(0.02)
∴PH = 1.7
B) according to this reaction:
by using ICE table:
HONH3+(aq) → H+(aq) + HONH2(aq)
intial 0.4 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka HONH3Cl = 9.09x10^-7
and Ka = [H+][HONH2] / [HONH3+]
So by substitution and we can assume [HONH3+] = 0.4 as the value of Ka is so small so,
9.09x10^-7 = X^2 / 0.4 by solving for X
∴ X = 6 x 10 ^-4
∴[H+] = 6x10^-4
PH = -㏒[H+]
= -㏒ (6x10^-4) = 3.22
when [H+] = 6x10^-4 + 0.02 m HCl
∴new value of PH = -㏒(6x10^-4+0.02)
= 1.69
C) when we have pure H2O and PH of water = 7
So we can get [H+] when PH = -㏒[H+]
∴[H+] = 10^-7 + 0.02MHCl
= 0.02
∴new value of PH = -㏒0.02
PH = 1.7
d) when HONH2 & HONH3Cl have the same concentration and Hcl added to them so we can assume that PH=Pka
and when we have Ka for HONH3Cl = 9.09x10^-7
So we can get the Pka:
Pka = -㏒Ka
= -㏒9.09x10^-7
= 6.04
∴PH = 6.04
and because of the concentration of the buffer components, HONH2 & HONH3Cl have 0.4 M and the adding of HCl = 0.02 M So PH will remain very near to 6
Answer:
The pH of the solution is 8.
Explanation:
To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:
1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:
pOH = –Log [OH-]
pOH = 6
6 = –Log [OH-]
–6 = Log [OH-]
[OH-] = Antilog (–6)
[OH-] = 1x10^–6 mol/L
2. The pH of the solution can be obtained as follow:
pH + pOH = 14
pOH = 6
pH + 6 = 14
pH = 14 – 6
pH = 8.
From the calculations made above,
[OH-] = 1x10^–6 mol/L
pH = 8.
Therefore, the correct answer is:
The pH of the solution is 8
Answer:
Explanation:
- State of benzene at RTP = liquid
- State of chloroform at RTP = liquid
- Boiling point of benzene = 80.1 °C
- Boiling point of chloroform = 61.2 °C
Since, both of the chemicals are liquids, we can separate it by the process of distillation.
<u>Distillation:</u>
- is the process in which we separate two liquids on the basis of their difference in boiling points.
<u>How it works:</u>
Since chloroform has less boiling point, it will evaporate and collected first and benzene will follow it after sometime.
- Apparatus of distillation is in the attached file.
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