A) Which of the following pairs cannot be mixed together to form a buffer soluti

Question

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?

Explanation / Answer

A.

When you add KOH to HF (weak acid), you get F-, the conjugated base of HF acid. Now, in the solution you have both HF and F-, which is a buffer.

When you add HCl to CH3COONa, you get CH3COOH (conjugated weak acid). Now, in the solution you have both CH3COONa and CH3COOH, which is a buffer.

B.

Since, HCN is a weak acid, it will dissociate partially. You won’t get large amount of [H+].

C.

Using Hendeson-Hesselbalach equation

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

or, pH =pKa + log {[A-] / [HA]}

or, 3.35 =pKa + log (0.150 / 0.200)

or, 3.35 =pKa + log (0.75)

or, 3.35 =pKa - 0.12

pKa = 3.35 + 0.12

pKa = 3.47

0.200 M of acid, HA in 0.5 L
Moles of HA = 0.2 M x 0.5 L = 0.1 mol

0.150 M of acid, HA in 0.5 L
Moles of HA = 0.2 M x 0.5 L = 0.075 mol

When you add 0.015 mol of NaOH, it will react with 0.015 mol of HA to form 0.015 mol of A-.

So, moles after addition of NaOH are
Moles of HA = 0.100 mol – 0.015 mol = 0.085 mol
Moles of A- = 0.075 mol + 0.015 mol = 0.085 mol

Total volume = 0.5 L

Concentration terms are
[HA] = 0.085 mol / 0.5 L = 0.17 M
[A-] = 0.085 mol / 0.5 L = 0.17 M

Again,

Using Hendeson-Hesselbalach equation

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

or, pH = pKa + log {[A-] / [HA]}

or, pH = 3.47 + log (0.17 / 0.17)

or, pH = 3.47 + log (1)

or, pH = 3.47 - 0

or, pH = 3.47

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