Thank you Trial 1 742 Trial 2 742 THE EQUILIBRIUM CONSTANT O

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Trial 1 7.42 Trial 2 7.42 THE EQUILIBRIUM CONSTANT OF A WEAK BASE A. pH of 0.50 M Unknown Weak Base pOH a. [OH] b. [HB\'] c. [B] 6.58 6.58

Solution

Answer

a) [OH-] = 2.63×10^-7M

b) [HB+] = 2.63×10^-7M

c) [B] ~ 0.50M

d) Keq = 1.38×10^-13M

Explanation

if B is the weak base, the dissociation of B is given as follows

B(aq) + H2O(l) <- - - - - > BH+(aq) + OH-(aq)

Keq = [BH+] [OH-] /[B]

at equillibrium,

[BH+] = x

[OH-] = x

[B] = 0.50 - x

Keq= x²/0.50 - x

from pH measurement, we know that pOH =6.58

pOH = - log[OH-]

-log[OH-] = 6.58

[OH-] = 1×10^-6.58 = 2.63×10^-7M

So, x = 2.63×10^-7M

So, at equillibrium

[OH-] = 2.63×10^-7M

[BH+] = 2.63×10^-7M

[B] = 0.50M - 0.000000263M ~ 0.50M

Applying these values

Keq =( 2.63×10^-7)²/0.50 = 1.38×10^-13