Question

The dissociation constant of dimethyl amine is 5.4 x 10^-3 at 298 K. What will be the pH of its 0.25 M solution and [H₂O] ion concentration ?

Answer 1

K_b = 5.4 x 10^-3

K_b = \(\frac{x^2}{0.25-x}\)

5.4 x 10^-3 = \(\frac{x^2}{0.25-x}\)

⇒ 0.00135 - .0..54x = x²

⇒ x² + 0.0054x - 0.00135 = 0

x = \(\frac{-0.0054\pm\sqrt{(0.0054)^2-(4\times(-0.00135))}}2\) 

x = 0.034

(taking only positive value of x)

So, concentration of \(\overline{O}H\) = 0.034

\(\therefore\) P^OH = -log[\(\overline{O}H\)]

 = -log [0.034]

P^OH = 1.47

\(\because\) P^H + P^OH = 14

\(\therefore\) P^H = 14 - 1.47

P^H = 12.53

\(\because\) [\(H_3O^+\)]  [\(\overline{O}H\)] = 1 x 10^-14

[\(H_3O^+\)] = \(\frac{1\times10^{-14}}{0.034}\)

[\(H_3O^+\)] = 2.94 x 10^-13