Question

A 25.0 mL solution of quinine was titrated with 0.50 M hydrochloric acid, HCl. It was found that the solution contained 0.100 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added

Answer 1

`pH=11.45`

Step-by-step explanation:

Hello there!

In this case, since the chemical equation representing the neutralization of the weak base quinine can be written as follows:

`C_(20)H_(24)N_2O_2+HCl\rightarrow C_(20)H_(25)N_2O_2^+Cl^-`

Whereas we have 0.100 moles of the base and those of acid as shown below

`n_(acid)=0.50molHCl/L*0.05000L=0.025molHCl`

Which means that the remaining moles of quinine are:

`n_{C_(20)H_(24)N_2O_2}^(final)=0.100mol-0.025mol=0.075mol`

And the resulting concentration:

`[C_(20)H_(24)N_2O_2]=(0.075mol)/((0.025+0.050)L) =1.00M`

Now, the calculation of the pH requires the pKb of quinine (5.1) as its ionization in water produces OH- ions:

`C_(20)H_(24)N_2O_2+H_2O\rightleftharpoons OH^-++C_(20)H_(25)N_2O_2^+`

And the equilibrium expression is:

`Kb=([C_(20)H_(24)N_2O_2][C_(20)H_(25)N_2O_2 ^+])/([C_(20)H_(24)N_2O_2]) \\\\10^(-5.1)=(x^2)/(1.00M)\\\\ 7.94x10^(-6)=(x^2)/(1.00M)`

Which is solved for x as follows:

`x= \sqrt{7.94x10^(-6)*1.00M}\\\\x=0.00282M`

Which means the pOH is:

`pOH=-log(0.00282)=2.55`

And the pH:

`pH=14-2.55\\\\pH=11.45`

Regards!