Which of the following will produce a buffered solution? For HCNO(aq), Ka = 2.2
×
10–4.

Question

Which of the following will produce a buffered solution? For HCNO(aq), Ka = 2.2 
×
 10–4.

BlackTom AI · Accepted Answer

Let’s break down the problem by considering what a buffer requires and then evaluate each choice step by step.
Option 1: 10 mL of 0.10 M NaCNO and 5.0 mL of 0.10 M HCl
- NaCNO provides CN-, the conjugate base of the weak acid HCNO.
- HCl is a strong acid that will protonate CN- to form HCNO until one reactant is used up.
- Moles: CN- from NaCNO = 0.010 L × 0.10 M = 0.0010 mol. HCl moles = 0.005 L × 0.10 M = 0.0005 mol.
- Reaction: CN- + H+ → HCNO proceeds for 0.0005 mol, leaving 0.0010 − 0.0005 = 0.0005 mol CN- and forming 0.0005 mol HCNO.
- The resulting mixture contains equal amounts (0.0005 mol each) of HCNO (weak acid) and CN- (its conjugate base), in a total volume of 15 mL. This is the classic buffer scenario, where the weak acid and its conjugate base coexist to resist pH changes.
- Therefore, this option can produce a buffered solution.
Option 2: 10 mL of 0.10 M NaCNO and 5.0 mL of 0.10 M NaOH
- NaOH provides OH–, which will not generate HCNO; instead, OH– would react with any HCNO present and/or leave just CN- as the dominant species.
- We begin with CN- but add base, which would shift the equilibrium away from the acid form and toward more CN- only; there would be little to no HCNO formed to establish a buffer system.
- Without a significant amount of both acid and conjugate base, buffering is not achieved.
Option 3: 10 mL of 0.10 M HCNO and 5.0 mL of 0.10 M HCl
- HCNO is the weak acid; adding HCl (a strong acid) will suppress its dissociation but does not introduce its conjugate base in appreciable amount.
- The mixture will be dominated by HCNO with extra H+, and there is not a substantial presence of CN- to form a conjugate-base pair.
- A buffer requires both HCNO and CN-; this setup lacks the conjugate base in useful quantity and does not resist pH changes in the typical buffer sense.
Option 4: 10 mL of 0.10 M NaCNO and 15 mL of 0.10 M HCl
- CN- from NaCNO will be protonated by H+ to form HCNO, possibly completely converting CN- to HCNO if HCl is in excess.
- Since more HCl is present than CN- (moles: CN- = 0.0010 mol; H+ = 0.015 L × 0.10 M = 0.0015 mol), the reaction consumes all CN- and leaves extra H+.
- The final mixture contains HCNO but little to no CN- to form a conjugate-base partner, so it does not constitute a buffer.
Option 5: 10 mL of 0.10 M HCNO and 15 mL of 0.10 M NaOH
- NaOH will deprotonate HCNO to form CN-; any HCNO present will be converted, increasing CN- concentration.
- The mixture ends up with CN- and possibly leftover OH–, but there is no weak acid present to pair with CN- in appreciable amounts to act as a buffer.
- Without the acid–base pair in the right proportion, buffering does not occur.
In summary, the only scenario that creates a mixture containing both HCNO and CN- in meaningful, roughly equal amounts (the classic buffer condition) is Option 1. The other options fail to establish the necessary acid–conjugate base pair in the right proportions or rely on excess strong species that prevent buffering behavior.

Which of the following will produce a buffered solution? For HCNO(aq), Ka = 2.2 × 10–4.Single choice

Similar Questions

Which pair does NOT form a buffer?

If a small amount of a strong base is added to a buffer made up of a weak acid, HA, and the sodium salt of its conjugate base, NaA, the pH of the buffer solution does not change appreciably because

Question 30 Choose a, b, c or d as the best answer. The author’s main argument is that the 100:80:100 model __________.

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler