(CH3)2NH(aq) + HNO3(aq) ---------------> (CH3)2NH2NO3
1 mole 1 mole
(CH3)2NH HNO3
M1 = 0.493M M2 = 0.242M
V1 = 170ml V2 =
n1 =1 n2 = 1
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 0.493*170*1/0.242*1 =346.32ml
Total volume of solution = 170 + 346.32 = 516.32ml = 0.51632L
no of moles of (CH3)2NH = molarity * volume in L
= 0.493*0.17 = 0.08381 moles
(CH3)2NH(aq) + HNO3(aq) ---------------> (CH3)2NH2NO3
I 0.08381moles 0.08381 moles 0
C -.0.08381 -0.08381 0.08381
E 0 0 0.08381
molarity of (CH3)2NH2NO3 = no of moles/volume in L
= 0.08381/0.51632 = 0.162M
PKb = 3.27
PKa = 14-PKb
= 14-3.27
PKa = 10.73
-logKa = 10.73
Ka = 10^-10.73 = 1.86*10^-11
(CH3)2NH2^+ (aq) + H2O (l) ----------------> (CH3)2NH(aq) + H3O^+ (aq)
I 0.162 0 0
C -x +x +x
E 0.162-x +x +x
Ka = [(CH3)2NH][H3O^+]/[(CH3)2NH2^+]
1.86*10^-11 = x*x/0.162-x
1.86*10^-11*(0.162-x) = x^2
x = 1.735*10^-6
[H3O^+] =x = 1.735*10^-6M
PH = -log[H3O^+]
= -log1.735*10^-6
= 5.76>>>>>answer
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