Answer:
[tex]pH=11.45[/tex]
Explanation:
Hello there!
In this case, since the chemical equation representing the neutralization of the weak base quinine can be written as follows:
[tex]C_{20}H_{24}N_2O_2+HCl\rightarrow C_{20}H_{25}N_2O_2^+Cl^-[/tex]
Whereas we have 0.100 moles of the base and those of acid as shown below
[tex]n_{acid}=0.50molHCl/L*0.05000L=0.025molHCl[/tex]
Which means that the remaining moles of quinine are:
[tex]n_{C_{20}H_{24}N_2O_2}^{final}=0.100mol-0.025mol=0.075mol[/tex]
And the resulting concentration:
[tex][C_{20}H_{24}N_2O_2]=\frac{0.075mol}{(0.025+0.050)L} =1.00M[/tex]
Now, the calculation of the pH requires the pKb of quinine (5.1) as its ionization in water produces OH- ions:
[tex]C_{20}H_{24}N_2O_2+H_2O\rightleftharpoons OH^-++C_{20}H_{25}N_2O_2^+[/tex]
And the equilibrium expression is:
[tex]Kb=\frac{[C_{20}H_{24}N_2O_2][C_{20}H_{25}N_2O_2 ^+]}{[C_{20}H_{24}N_2O_2]} \\\\10^{-5.1}=\frac{x^2}{1.00M}\\\\ 7.94x10^{-6}=\frac{x^2}{1.00M}[/tex]
Which is solved for x as follows:
[tex]x= \sqrt{7.94x10^{-6}*1.00M}\\\\x=0.00282M[/tex]
Which means the pOH is:
[tex]pOH=-log(0.00282)=2.55[/tex]
And the pH:
[tex]pH=14-2.55\\\\pH=11.45[/tex]
Regards!
