Assume K_w = 1.0 times 10^-14 for each problem unless otherwise stated. (a) What

Question

Assume K_w = 1.0 times 10^-14 for each problem unless otherwise stated. (a) What are the equilibrium concentrations of C_6H_5NH_2, C_6H_5NH^+_3, Cl^-, H^+, and OH^- in an aqueous solution prepared by combining 50.0 mL of 0.0400 M C_6H_5NH^+_3Cl^- and 50.0 mL of 0.0100 M KOH? K_b = 7.4 times 10^-10 for C_6H_5NH_2. (b) Is the resultant solution in Part a above a buffer solution? Justify your choice. (a) What are the equilibrium concentrations of NO^-_2, HNO_2, H^+ and OH^- in an aqueous solution prepared by combining 50.0 mL of 0.0500 M NaNO_2 and 50.0 mL of 0.0800 M HCl? K_a = 7.2 times 10^-4 for HNO_2. (b) Is the resultant solution in Part a above a buffer solution? Justify your choice.

Explanation / Answer

Question 1.

a)

Find equilibrium concentrations

note that

C6H5NH3Cl will act as acid --> C6H5NH3+ + Cl-

C6H5NH3+ + H2O <--< C6H5NH2 +´H3O+

Note that if there is KOH addition, then

C6H5NH3+ + OH- --> C6H5NH2 + H2O

so

mmol of acid = MV = 50*0.04 = 2

mmol of base = MV = 50*0.01 = 0.5

then, after reaction

mmol of R-NH3+ left = 2-0.5 = 1.5

mmol of R-NH2 formed = 0 + 0.5 = 0.5

This is a buffer

so

pH = pKa + log(R-NH3+ / R-NH2)

substitute data

pKa = 14-pKb = 14 + log(Kb) = 14 + log(7.4*10^-10) = 4.8692

pH = 4.8692+ log(1.5 / 0.5)

pH = 5.3463

[H+] = 10^-pH = 10^-5.3463 = 0.00000450505M

[OH-] = 10^-pOH = 10^-(14-5.3463) = 2.219*10^-9 M

C6H5NH2 = 0.5 mmol / (50+50) = 0.5/100 = 0.005 M

C6H5NH3+ = 1.5 mmol / (50+50) = 0.015 M

[C-] = 2 mmol / (100) = 0.020 M

b)

Yes, this is a buffer since there is weak acid, C6H5NH3+ and conjugate base formed, C6H5NH2

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