Question #167613

Trovare quanti ml di una soluzione di LiOH 0.8M si devono aggiungere a 120ml di una soluzione di metilammina,CH3NH2 0.4M sapendo che questa base ha Kb=4*10^-5 perché la concentrazione deglio ioni CH3NH3+ risulti 3*10^-5M


1
Expert's answer
2021-03-01T02:36:06-0500

Kb = [CH3NH3+][OH][CH3NH2]\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} = 4×\times10-5

n(CH3NH2) = 0.4×\times0.120 = 0.048 mol

n(OH-) = n(LiOH)

Let the volume of LiOH needed is x liters, then total volume is 0.120+x, Ctotal(CH3NH2) = 0.0480.120+x\frac{0.048}{0.120+x} , [OH-] = 0.8x0.12+x\frac{0.8x}{0.12+x} . As we know from the task [CH3NH3+] = 3×\times10-5, then [CH3NH2] = 0.048(3×105×(0.120+x))0.120+x\frac{0.048- (3\times10^{-5}\times(0.120+x))}{0.120+x} ; So the total equotion is:

3×105×0.8x0.12+x0.048(3×105×(0.120+x))0.120+x\frac{3\times10^{-5}\times\frac{0.8x}{0.12+x}}{\frac{0.048- (3\times10^{-5}\times(0.120+x))}{0.120+x}} = 4×\times10-5

Solving this equotion we get x = 0.08 L = 80 ml.

Answer: V(LiOH) = 80 ml.


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