Question
Which of the following will produce a buffer solution when mixed in equal volumes?
A.
$$0.1\,mol\,d{m^{ - 3}}\,N{H_4}OH\,\,{\text{and}}$$ $$0.1\,mol\,d{m^{ - 3}}\,HCl$$
B.
$$0.05\,mol\,d{m^{ - 3}}\,N{H_4}OH\,\,{\text{and}}$$ $$0.1\,mol\,d{m^{ - 3}}\,HCl$$
C.
$$0.1\,mol\,d{m^{ - 3}}\,N{H_4}OH\,\,{\text{and}}$$ $$0.05\,mol\,d{m^{ - 3}}\,HCl$$
D.
$$0.1\,mol\,d{m^{ - 3}}\,C{H_3}COONa\,\,{\text{and}}$$ $$0.1\,mol\,d{m^{ - 3}}\,NaOH$$
Answer :
$$0.1\,mol\,d{m^{ - 3}}\,N{H_4}OH\,\,{\text{and}}$$ $$0.05\,mol\,d{m^{ - 3}}\,HCl$$
Solution :
In $$0.1\,mol\,d{m^{ - 3}}\,N{H_4}OH$$ and $$0.05\,mol\,d{m^{ - 3}}\,HCl,$$ total amount of $$HCl$$ reacts with $$N{H_4}OH$$ to form $$N{H_4}Cl$$ and some $$N{H_4}OH$$ will be left unreacted. Thus, the resultant solution contains $$N{H_4}Cl$$ and $$N{H_4}OH$$ which will produce a buffer solution.