A) Which of the following pairs cannot be mixed together to form a buffer solution? Why?

A. NH3, NH4Cl

B. KOH, HF

C. NaC2H3O2, HCl

D. H3PO4, KH2PO4

E. RbOH, HCl

I already know the answer is E. I want help understanding when we can rule out choices B and C.

B) For a solution equimolar in HCN and NaCN, which statement is false?

Without giving all the choices, the false choice is "The [H+] is larger than it would be if only the HCN was in solution."

Why? How would I know this?

C) A buffer containing 0.200 M of acid, HA, and 0.150 M of its conjugate, A-, has a pH of 3.35. What is the pH after 0.015 mol of NaOH is added to 0.500 L of the solution.

How do I solve this?

Part 1)

A buffer is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa

A. NH₃, NH₄Cl

Here, NH₄Cl will dissociate in aqueous solution to form NH₄⁺ and Cl^-. Therefore, the solution contains the weak base NH₃ and its conjugated acid NH₄⁺. Thus, this combination can act as a buffer

B. KOH, HF

Here, KOH is a strong base and HF is a weak acid. KOH will consume a part of the H⁺ ions from HF. The resulting solution will have contained HF and F^- which can act as a buffer

C. NaC₂H₃O₂, HCl

Sodium acetate is a salt of acetic acid. In the presence of HCl, it forms acetic acid. Thus, the resulting solution will contain

sodium acetate and acetic acid which can act as a buffer

D. H₃PO₄, KH₂PO₄

Phosphoric acid is a weak acid and KH₂PO₄ is its potassium salt. This combination can act as a buffer

E. RbOH, HCl

RbOH is a strong base and HCl is a strong acid. This combination will not function as a buffer

Part 2)

HCN(aq)  H⁺(aq) + CN^-(aq)

Thus, if NaCN is also present in the system it’ll decrease the concentration of H⁺ by common ion effect

Therefore,

[H⁺] is larger than it would be if only the HCN were in solution

Part 3)

Moles of acid = 0.2 M x 0.5 L = 100 mmoles

Moles of salt = 0.15 M x 0.5 L = 0.075 mmoles

Applying the Henderson-Hesselbalach equation

pH =pKa + log[salt]/[acid]

pH = pKa + log (75/100)

pKa = 3.35 - log (75/100) = 3.35 + 0.125 = 3.475

pKa = 3.35 + 0.125 = 3.475

pKa = 3.475

After 0.015 mol of NaOH is added

0.015 moles = 15 mmoles

The 15 mmoles of the base will react with the 15 mmoles of the acid

This will increase the concentration of salt by 15 mmoles and decrease the concentration of acid by 15 mmoles

Therefore, the resulting pH can be calculated as,

pH = pKa + log (75+15) /(100-15)

pH = 3.475 + log(90/85) = 3.49

pH = 3.49