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 quini

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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

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Thiên Ân 2 weeks 2021-07-21T22:02:25+00:00 1 Answers 0 views 0

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    2021-07-21T22:03:40+00:00

    Answer:

    pH=11.45

    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]=\frac{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=\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}

    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!

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