Answer to Question #167613 in Organic Chemistry for Alessia

Question

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

Answer 1

K_b = $\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$ = 4$\times$10^-5

n(CH₃NH₂) = 0.4$\times$0.120 = 0.048 mol

n(OH^-) = n(LiOH)

Let the volume of LiOH needed is x liters, then total volume is 0.120+x, C_total(CH₃NH₂) = $\frac{0.048}{0.120+x}$ , [OH^-] = $\frac{0.8x}{0.12+x}$ . As we know from the task [CH₃NH₃⁺] = 3$\times$10^-5, then [CH₃NH₂] = $\frac{0.048- (3\times10^{-5}\times(0.120+x))}{0.120+x}$ ; So the total equotion is:

$\frac{3\times10^{-5}\times\frac{0.8x}{0.12+x}}{\frac{0.048- (3\times10^{-5}\times(0.120+x))}{0.120+x}}$ = 4$\times$10^-5

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

Answer: V(LiOH) = 80 ml.