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JEE Advanced 2017 Paper 1 · Chemistry · Ionic Equilibrium · medium

A buffer is prepared by mixing 50.0 mL of 0.200 M NH3 with 50.0 mL of 0.100 M HCl at 298 K. Take Kb(NH3)=1.8×10^-5, log 1.8=0.255, Kw=1.0×10^-14, and…

One or more options correct

A buffer is prepared by mixing 50.0 mL50.0\ \mathrm{mL} of 0.200 M0.200\ \mathrm{M} NH3\mathrm{NH_3} with 50.0 mL50.0\ \mathrm{mL} of 0.100 M0.100\ \mathrm{M} HCl\mathrm{HCl} at 298 K298\ \mathrm{K}. Take Kb(NH3)=1.8×10−5K_b(\mathrm{NH_3})=1.8\times10^{-5}, log⁡1.8=0.255\log 1.8=0.255, Kw=1.0×10−14K_w=1.0\times10^{-14}, and assume additive volumes. The CORRECT statement(s) is/are
  1. A. The pH\mathrm{pH} of the buffer is 9.269.26.
  2. B. Adding a further 5.0 mL5.0\ \mathrm{mL} of 0.100 M0.100\ \mathrm{M} HCl\mathrm{HCl} to this buffer lowers its pH\mathrm{pH} by less than 0.100.10 unit.
  3. C. If the buffer is titrated with 0.100 M0.100\ \mathrm{M} HCl\mathrm{HCl} until the total acid added (including the original 50.0 mL50.0\ \mathrm{mL}) is 100.0 mL100.0\ \mathrm{mL}, the pH\mathrm{pH} at that stage is 7.007.00 because the salt formed is neutral.
  4. D. A further 41 mL41\ \mathrm{mL} of 0.100 M0.100\ \mathrm{M} HCl\mathrm{HCl} (approximately) must be added to the original buffer to bring its pH\mathrm{pH} down to 8.268.26.
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Answer: A, B, D

Initially there are 10.010.0 mmol NH3\mathrm{NH_3} and 5.05.0 mmol HCl\mathrm{HCl}, leaving 5.05.0 mmol NH3\mathrm{NH_3} and 5.05.0 mmol NH4+\mathrm{NH_4^+}. Then pKa(NH4+)=14−pKb=14−4.745=9.255\mathrm{pK_a(NH_4^+)}=14-\mathrm{p}K_b=14-4.745=9.255 and pH=pKa+log⁡([NH3]/[NH4+])=9.26\mathrm{pH}=\mathrm{p}K_a+\log([\mathrm{NH_3}]/[\mathrm{NH_4^+}])=9.26, so (A) is correct. (B) 0.500.50 mmol HCl\mathrm{HCl} converts 0.500.50 mmol NH3\mathrm{NH_3} to NH4+\mathrm{NH_4^+}, so the ratio becomes 4.5/5.54.5/5.5 and ΔpH=log⁡(4.5/5.5)=−0.087\Delta\mathrm{pH}=\log(4.5/5.5)=-0.087, a fall of less than 0.100.10: correct. (C) 100.0 mL100.0\ \mathrm{mL} of acid supplies 10.010.0 mmol, exactly neutralising the 10.010.0 mmol NH3\mathrm{NH_3} (equivalence point). The solution contains only NH4+\mathrm{NH_4^+} (0.0667 M0.0667\ \mathrm{M} in 150 mL150\ \mathrm{mL}), a weak acid, so pH=12(pKa−log⁡C)=5.22\mathrm{pH}=\tfrac12(\mathrm{p}K_a-\log C)=5.22, not 7.007.00: wrong. (D) For pH=8.26\mathrm{pH}=8.26 the ratio [NH3]/[NH4+]=10−1[\mathrm{NH_3}]/[\mathrm{NH_4^+}]=10^{-1}, so NH4+=10×(10/11)=9.09\mathrm{NH_4^+}=10\times(10/11)=9.09 mmol; the additional acid is 9.09−5.0=4.099.09-5.0=4.09 mmol, i.e. 40.9≈41 mL40.9\approx41\ \mathrm{mL}: correct.
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