Question

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was the final pH at the equivalence point

Answer 1

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

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