how much Ch3Nh2 kb=10-4 should be dissolved in 5L of water such that final solution contains 5×10-11 moles of hydrogen ion
Answers
Explanation:
The reaction occuring is :
0.02
0.1
CH
3
NH
2
+
0
0.08
HCl
→
0.08
0
CH
3
NH
3
+
Cl
Thus, the reaction mixture contains unreacted methylamine and its hydrochloride salt.
Thus, it is a basic buffer solution.
The expression for the hydroxide ion concentration is [OH
−
]=K
b
×
[Conjugate acid]
[Base]
Substitute values in the above expression.
[OH
−
]=5×10
−4
×
0.08
0.02
=1.25×10
−4
But [H
+
][OH
−
]=k
w
=10
−14
∴[H
+
]=
[OH
−
]
10
−14
=
1.25×10
−4
10
−14
=8×10
−11
M
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Answer:
1.55g of CH3NH2
Explanation:
Given
Kb = 10^-4
5 liters of water
final solution contains 5 × 10^-11 moles of H+ ions
wt of CH3NH2 is 31
CH3NH2 <==> CH3NH3^+ + OH^-
[H+] = n/v = 5 × 10^-11/5
[H+] = 10^-11
pH of H+ ion is 11
pOH = 3
[OH-] = 10^-3
Kb = [CH3NH3+][OH-] / [CH3NH2]
Kb = C{`alpha`}
{`alpha`} = 10^-1
[OH-] = C{`alpha`} = 10^-3
C = 10^-3 / 10^-1
C = 10^-2 = n / 5
n = 5 × 10^-2 = wt / 31
wt = 1.55g