Buffer Preparation

Jay Shniderson

Experiment 8: Buffer Preparation

Analytical Chemistry Lab CHEM 2120.03

Dr. Richard Quy

11 March 2015

Introduction

        The purpose of Experiment 8: Preparation of a Buffer Solution was to create a buffer with the listed materials that matched the individual’s given pH. The correct solutions that matched up with the given pH were chosen based on the closest pK values that were recorded in the pre-lab table. The required calculations included the relative concentrations of each component, followed by the preparation of s solution containing both ingredients using two different methods. The first method was to prepare a mixture from a solution or solid form of the

acid and its conjugate salt; the second was to prepare a mixture from a solution of the acid or base and adding HCl or NaOH to prepare its conjugate in situ.

        I was given the pH of 9.3, which correlated to the pK values of ammonium hydroxide (aq) and ammonium chloride (s). Because the concentration of my base, ammonium hydroxide, was greater than my acid, ammonium chloride, I used HCl for my second method of creating the buffer solution.

        Buffers are useful in chemistry because they can resist pH chance after the addition of an acid or base. Also, it can neutralize small amounts of added acid or base in a solution which will help the system’s pH remain somewhat stable. Composed of a weak conjugate acid/base pairing, a buffer’s purpose is to maintain the pH range of a mixture of a stronger acid/base solution pairing.

Data and Results

        pH of my given solution= 9.3

        Acid= ammonium hydroxide, conjugate base= ammonium chloride (pK= 9.25)

Calculation: Henderson- Hasselbalch equasion

        pH = pKa + log ([base] / [acid])

        9.3 = 9.25 + log ([base] / [acid])

        .05 = log ([base] / [acid])

        10^.05 = ([base] / [acid])

        1.122 = ([base] / [acid])

        1.122 = ([x] / [.05 - x])

        .056 - 1.122x = x

        .056 = 2.122x —> x = .0264 [Base]

        .05 - x = .05 - .0264 = .0236 [Acid]

Volume needed for first method:

Base: ( .0264 mmol / mL ) (100 mL / 1 L ) = 2.62 mL

Acid: ( .0236 mmol / mL ) ( 53.49 g / mol ) = 1.262 g

Resulting pH from the mixture: 8.54

Method 2: equasion NH4OH + HCl —> H2O + NH4Cl

        pH = pKa + log ([base] / [acid])

        9.3 = 9.25 + log ([base] / [acid])

        .05 = log ([base] / [acid])

        10^.05 = ([base] / [acid])

        1.122 = ([base] / [acid])

        1.122 = ([x] / [.05 - x])

        .056 - 1.122x = x

        .056 = 2.122x —> x = .0264 [Base]

        .05 - x = .05 - .0264 = .0236 [Acid] (use HCl instead of ammonium chloride)