Question

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if thepH of the solution was 2.62?A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate the Kb for ethylamine.

Answer 1

Answer:

$Ka=0.000346$

$Kb=0.000593$

Explanation:

We have to start with the calculation of the concentration of the acid, using the molar mass:

Molas mass of $C_9H_8O_4$= 180.16 g/mol

$2*g\frac{1~mol}{180.16~g}=~0.011~mol$

$M=\frac{0.011~mol}{0.6~L}=0.019~M$

The,  with the ionization equation for aspirin, using the ICE table, we can calculate the Ka expression.

            HA     <=>    A^-      +      H^+

I       0.019             Zero            Zero

C           -X               +X               +X

E     0.019-X              X                  X

$Ka=\frac{[X][X]}{[0.019-X]}=\frac{[X]^2}{[0.019-X]}$

Now, we can calculate X using the pH value:

$pH=-Log[H^+]$

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^6^2=0.00240$

With this value we can calculate the Ka, so:

$Ka=\frac{[0.00240]^2}{[0.019-0.00240]}=0.000346$.

For the ethylamine we have to follow the same procedure:

         B^-    +    H2O     <=>    BH      +      OH^-

I          0.1              /                  Zero            Zero

C           -X            /                   +X               +X

E          0.1-X          /                    X                  X

$Kb=\frac{[X][X]}{[0.1-X]}=\frac{[X]^2}{[0.1-X]}$

Now, we can calculate X using the pH value:

$14=pH+pOH$

$pOH=14-pH$

$pOH=-Log[OH^-]$

$[OH^-]=10^-^p^O^H$

$[OH^-]=10^-^2^.^1^3=0.00741$

With this value we can calculate the Kb, so:

$Kb=\frac{[0.00741]^2}{[0.1-0.00741]}=0.000593$