Question

The base protonation constant Kb of allantoin (C4H4N3O3NH2) is ×9.1210−6. Calculate the pH of a 1.1M solution of allantoin at 25°C. Round your answer to 1 decimal pl

Answer 1

Answer:

pH= 11.5

Explanation:

To the calculation of the pH value, we need to write the ionization equation first. Then using the ICE table and the information given by the problem we can set up the problem:

$C_4H_4N_3O_3NH_2~+~H_2O->C_4H_4N_3O_3NH_3^+~+~OH^-$

I                  1.1 M                                                Zero                             Zero              

C                   -X                                                   +X                                 +X

E                  1.1-X                                                   X                                   X

As next step, using the equilibrium expression for Kb we can replace the terms find in the ICE table and then solve for X, so:

$Kb=\frac{[C_4H_4N_2O_3NH_3^+][OH^-]}{[C_4H_4N_3O_3NH_2]}$

$9.2x10^-^6=\frac{[X][X]}{[1.1-X]}$

$9.2x10^-^6=\frac{X^2}{1.1-X}$

$(9.2x10^-^6)*[1.1-X]=X^2$

$1.01x10^-^5~-9.2x10^-^6X-X^2=0$

$X=0.003173$

With the value of X we can find the pOH (lets remember that X is the concentration of OH-), so:

$pOH=-Log(0.003173)$

$pOH=2.50$

Finally, we can find the pH value with the equation:

$14~=~pH~+~pOH$

$14~=~2.5~+~pOH$

$pH=14-2.5$

$pH=11.5$

The 1.1 M allantoin solution have a pH value of 11.5