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Chapter 10 pH, buffers, and buffering.

OBJECTIVE

In this chapter you will learn

* definitions of acids and bases.

* pH calculations.

* differences between normality and molarity.

* how to neutralize strong acids and bases.

* calculations for titrating weak acids and bases.

* how to create buffers of varying strength.

Overview

Buffers are critical in soil science because so much of the chemical and biological response of the soil environment is influenced by soil pH and its fluctuations. Buffers do exactly what their name implies. They buffer, or protect, the environment from sudden change. In this chapter you will look at some calculations you'll need to know to work with pH and buffers in soil science.

Acids, Bases, and Water

The Bronsted definition of an acid is a substance that donates protons ([H.sup.+]) and a base as a substance that accepts [H.sup.+].

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII [10-1]

A strong acid completely ionizes to yield the corresponding acid and base pairs. For example, HCl is a strong acid because in water:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII [10-2]

The hydronium ion ([H.sub.3][O.sup.+]) and [H.sup.+] are identical for all intents and purposes.

Likewise, a strong base is a compound that also ionizes completely in water:

NaOH [right arrow] [Na.sup.+] + O[H.sup.-] [10-3]

Water can dissociate (come apart) to form [H.sup.+] and O[H.sup.-]. In pure water, the [H.sup.+] concentration is [10.sup.-7] M. For every mole of [H.sup.+] produced, 1 mol of O[H.sup.-] is produced, and thus the OH- concentration is [10.sup.-7]. Remember that 1 mol of any substance is 6.023 x [10.sup.23] molecules of that substance, and that the concentration of most substances in water is quantified in terms of molarity (M) or moles per liter (mol [L.sup.-1]).

pH

The shorthand notation for describing how much [H.sup.+] is in water is called pH. pH stands for the negative logarithm of the hydrogen-ion concentration (abbreviated as [[H.sup.+]]) and is expressed as:

pH = -log[[H.sup.+]] [10-4]

pOH stands for the negative logarithm of the hydroxyl concentration:

pOH = -log[O[H.sup.-] [10-5]

Because pH + pOH = 14, if you know pH, you also know pOH. The important point to remember is that the concentration of [H.sup.+] increases as the pH declines, and decreases as the pH rises.

Example 10-1

What is the pH of a 0.002 M solution of HCl?

Solution

HCl is a strong acid that completely dissociates. So, the [[H.sup.+]] will be nearly 0.002 M.

pH = -log[0.002 M] = 2.7

Example 10-2

What is the pH of a [10.sup.-8] M solution of HCl?

Solution

-log[[10.sup.-8] M] = 8, but the pH [not equal to] 8. Remember that the ionization of pure water gives a [[H.sup.+]] of [10.sup.-7] M. So, the pH would be -log ([10.sup.-7] + [10.sup.- 8]) = 6.96

Molarity and Normality

You have been dealing with problems that look at the concentration of [H.sup.+] in terms of its molarity (M = moles/liter). An important concept in buffering and acid/base chemistry is normality (N = equivalents/ liter). Just as moles = grams/molecular weight, equivalents = grams/ equivalent weight. Equivalent weight and molecular weight are related by the equation:

EW = MW/n [10-6]

where

EW = equivalent weight

MW = molecular weight

n = number of [H.sup.+] or O[H.sup.-] that can be replaced if you're dealing with acids and bases, or number of electrons gained or lost per molecule if you're dealing with oxidation and reduction reactions or molecular charge

Another way of expressing this is as:

Equivalents = n x moles

Molarity and normality are related by the equation:

N = nM [10-7]

Example 10-3

How many moles or equivalents are in 28 g NaOH?

Solution

The molecular weight of NaOH is 40 g [mol.sup.-1].

The number of moles = 28 g/40 g [mol.sup.-1] = 0.7 mol

The equivalent weight of NaOH = molecular weight/ n

Only one O[H.sup.-] is present in NaOH, so n = 1

Therefore, there are also 0.7 eq in 28 g of NaOH

Example 10-4

What is the normality of a 0.03 M solution of [H.sub.2] S[O.sub.4]?

Solution

There are two ionizable [H.sup.+] associated with [H.sub.2] S[O.sub.4], so n = 2.

N = nM, so N = (2)(0.03 M)

N = 0.06

Example 10-5

How many moles and equivalents of [Ca.sup.2+] are in 50 g of Ca[Cl.sub.2]?

Solution

There are 111 g Ca[Cl.sub.2]/mol Ca[Cl.sub.2].

(50 g Ca[Cl.sub.2]) x mol Ca[Cl.sub.2] / 111 g Ca[Cl.sub.2] = 0.45 mol Ca[Cl.sub.2]

There is one atom of Ca in Ca[Cl.sub.2] so there are also 0.45 mol Ca.

Equivalents = n x moles. Since the charge on Ca = 2+, n = 2.

So, 0.45 mol Ca x 2 = 0.90 eq Ca.

Example 10-6

What is the normality of a 0.05 M solution of Ca[Cl.sub.2]?

Solution

The charge on Ca is 2+, so n = 2.

N = nM, so N = (2)(0.05 M)

N = 0.10 Ca[Cl.sub.2]

Neutralizing Strong Acids and Bases

One of the major procedures for assessing C[O.sub.2] evolution from soil is to trap the C[O.sub.2] in a strong base solution such as 1 M NaOH and subsequently titrating the solution with a strong acid such as HCI. Titration stops when all the remaining O[H.sup.-] has been neutralized. This also becomes important in considering liming reactions to raise soil pH. So, it is important to understand neutralization reactions quantitatively.

The total equivalents of acid required to neutralize the total equivalents of base are exactly equal to one another. You can relate the two by a simple equation:

([N.sub.acid])([Volume.sub.acid]) = ([N.sub.base])([Volume.sub.base]) [10-8]

Example 10-7

How many milliliters of 0.3 M [H.sub.2] S[O.sub.4] are required to exactly neutralize 250 mL of 0.75 M KOH?

Solution

The first thing to do in working neutralization reactions is to put everything on the same equivalent basis. 0.75 M KOH = 0.75 N KOH because normality and molarity are identical for compounds that dissociate to yield ions with a single charge. On the other hand, the normality of 0.3 M [H.sub.2] S[O.sub.4] is 0.6 because N = nM and [H.sub.2] S[O.sub.4] has two protons to exchange. To answer the question, you write the equation:

(N [H.sub.2] S[O.sub.4])(Volume [H.sub.2] S[O.sub.4]) = (N KOH)(Volume KOH)

(0.6 N [H.sub.2] S[O.sub.4])(mL [H.sub.2]S[O.sub.4]) = (0.75 N KOH)(250 mL KOH) = 312.5 mL [H.sub.2] S[O.sub.4]

Titrating Weak Acids--The Henderson--Hasselbach Equation

Strong acids and bases completely dissociate when they are added to water, but weak acids like acetic acid (C[H.sub.3]COOH), for example, do not (Segal, 1976). Weak acids form equilibrium relationships in water:

HA [left and right arrow] [H.sup.+] + [A.sup.-] [10-9]

where HA is the undissociated acid, and [H.sup.+] and [A.sup.-] represent the acid and base components of the acid, respectively.

The extent to which the weak acid dissociates is described in Equation 10-10:

[K.sub.a] = [[H.sup.+]][[A.sup.-]]/[HA] [10-10]

where [K.sub.a] is the dissociation constant of the weak acid and is usually a very small number. Like pH, it is frequently reported as [pK.sub.a] (the negative logarithm of the dissociation constant [K.sub.a]). The pH of a weak acid is described by the equation:

pH = ([pK.sub.a] + pHA) / 2 [10-11]

Example 10-8

What is the pH of a 0.15 M solution of a weak acid (HA) that has a [K.sub.a] of [10.sup.-5] M?

Solution

The [pK.sub.a] of [10.sup.-5] M = 5

pHA = -log[0.15 M] = 0.82

pH = ([pK.sub.a] + pHA) / 2

pH = (5 + 0.082) / 2 = 2.54

Example 10-9

What is the pH of a 10 mM solution of N[H.sub.4.sup.+] (a weak acid), which has a [pK.sub.a] of 9.2?

Solution

pHA = -log (0.01 M N[H.sub.4.sup.+]) = 2

pH = ([pK.sub.a] + pHA) / 2

pH = (9.2 + 2.0) / 2 = 5.6

Values for [pK.sub.a] of various buffering systems are listed in Appendix 1.

For a weak acid, the pH at any time can be described by the Henderson-Hasselbach equation:

pH = p[K.sub.a] + log [[A.sup.-] / [HA] [10-12]

When the weak acid solution is exactly half titrated, so that [HA] = [[A.sup.-]], the pH = p[K.sub.a].

Example 10-10

What is the pH of 400 mL of weak acid (0.15 M) titrated with 0.2 M KOH, after 50 mL of KOH has been added? Assume the p[K.sub.a] of the acid is 5.5.

Solution

50 mL = 0.05 L KOH

(0.05 L KOH) x (0.2 M KOH) = 0.01 mol KOH = 0.01 eq KOH

(0.15 M HA) x (0.4 L HA) = 0.06 mol HA = 0.06 eq HA

0.06 mol - 0.01 mol = 0.05 mol HA left

pH = p[K.sub.a] + log [[A.sup.-]] / [HA]

pH = 5.5 + log (0.01 mol [A.sup.-]) / (0.05 mol HA) = 4.8

Buffers and Buffering

Common buffers are mixtures of a conjugate acid and base (C[H.sub.3]COOH and C[H.sub.3]CO[O.sup.-], for example). Mixtures of these weak acids or bases resist pH change by accepting [H.sup.+] or O[H.sup.-] added to the system. If [H.sup.+] is added, for example, C[H.sub.3]CO[O.sup.-] will convert to C[H.sub.3] COOH, while if O[H.sup.-] is added, C[H.sub.3] COOH will further dissociate to release [H.sup.+] and subsequently form [H.sub.2]O.

p[K.sub.a] is important in buffered systems (hence, the reason you have looked at it in detail) because the ability of a buffer to resist changes in pH is greatest around the p[K.sub.a]. If you want a buffer that will strongly resist pH changes and keep a solution of approximately pH 7, you choose a buffer with a p[K.sub.a] of approximately 7 rather than one with a p[K.sub.a] of 5.5.

Example 10-11

How much C[H.sub.3]COOH and C[H.sub.3] CO[O.sup.-] are in a 0.25 M buffer that has a pH of 6.0 given that the [K.sub.a] for acetate buffers is 1.7 x [10.sup.-5]?

Solution

p[K.sub.a] = -log [1.7 x [10.sup.-5]] = 4.77

pH = p[K.sub.a] + log [[A.sup.-]] / [HA]

6 = 4.77 + log ([C[H.sub.3] CO[O.sup.-] / [C[H.sub.3]COOH])

If the concentration of the buffer is 0.25 M, and the total concentration of C[H.sub.3] COOH and C[H.sub.3] CO[O.sup.-] = 0.25 M, you can then solve for the concentration of either C[H.sub.3] CO[O.sup.-] or C[H.sub.3]COOH algebraically.

0.25 M = [C[H.sub.3] COOH] + [C[H.sub.3] CO[O.sup.-]]

0.25 M - [C[H.sub.3] COOH] = [C[H.sub.3] CO[O.sup.-]]

6 = 4.77 + log (0.25 M - ([C[H.sub.3]COOH] / [C[H.sub.3]COOH])

1.23 = log (0.25 M - [C[H.sub.3]COOH] / [C[H.sub.3]COOH])

[10.sup.1.23] = (0.25 M - [C[H.sub.3] COOH] / [C[H.sub.3]COOH])

16.98 = (0.25 M - [C[H.sub.3] COOH] / [C[H.sub.3]COOH])

16.98 [C[H.sub.3] COOH] = (0.25 M - [C[H.sub.3]COOH])

16.98 [C[H.sub.3] COOH] + [C[H.sub.3] COOH] = 0.25 M

17.98 [C[H.sub.3] COOH] = 0.25 M

[C[H.sub.3] COOH] = 0.25 M / 17.98 = 0.014 M

If [C[H.sub.3] COOH] = 0.014 M, then [C[H.sub.3] CO[O.sup.-]] = 0.236 M

6 = 4.77 + log 0.236 / 0.014

Most effective environmental buffers have a p[K.sub.a] ranging from 3 to 9. Preparing buffers for laboratory use and chemical analysis is an important activity in soil science.

Example 10-12

How do you prepare 2 L of 0.3 M acetate buffer at pH 5.0 if your starting ingredients are sodium acetate trihydrate (NaC[H.sub.3]CO[O.sup.-].3[H.sub.2] O, MW = 136) and 2 M acetic acid (C[H.sub.3] COOH)?

Solution

pH = p[K.sub.a] + log [[A.sup.-]] / [HA]

5.0 = 4.7 + log [[A.sup.-]] / [HA]

[A.sup.-] + HA = 0.3 M

[A.sup.-] = 0.3M - HA

0.3 = log [0.3 M - HA] / [HA]

1.995 HA = 0.3 M - HA

2.995 HA =0.3M

HA = 0.10 M

[A.sup.-] = 0.20 M

The HA comes from the acetic acid. To calculate how much you need:

(2 L) x (0.10 mol [L.sup.-1] HA) = 0.20 mol HA

0.20 mol HA/2 mol [L.sup.-1] acetic acid = 0.10 L of acetic acid.

The [A.sup.-] comes from the sodium acetate trihydrate, which has a molecular weight of 136 g [mol.sup.-1].

(0.20 mol [L.sup.-1] [A.sup.-]) x 2 L = 0.40 mol [A.sup.-] x 136 g mol

= 54.4 g sodium acetate trihydrate

So, to make the buffer, dissolve 54.4 g of sodium acetate trihydrate in some water, add 100 mL of 2 M acetic acid, and dilute to 2 L, and write down the amounts you need so you don't have to go through all the calculations the next time you make the buffer.

An easier approach to buffer preparation is to realize that when the acid and base conjugates of a buffer system are in equimolar concentrations, pH = p[K.sub.a]. If you want to raise the pH by one unit from the p[K.sub.a], there should be 10 times as much of the acid form as the base form. A useful observation is that if you make a dilute solution with only the acid form of the compound, the pH will be approximately two units less than the p[K.sub.a]. Conversely, if you make a dilute solution with only the base form of a compound, the p[K.sub.a] will be approximately two units greater than the p[K.sub.a].

Example 10-13

Prepare 1 L of a 0.2 M phosphate buffer at pH 7.2.

Solution

The [H.sub.2] P[O.sub.4.sup.-] [left and right arrow] [H.sup.+] + HP[O.sub.4.sup.-] buffer system has a p[K.sub.a] of 7.2. To obtain a pH 7.2 buffer, you simply have to have equimolar amounts of [H.sub.2] P[O.sub.4.sup.-] and HP[O.sub.4.sup.2-] with a total molar concentration of 0.2 M. In other words [[H.sub.2] P[O.sub.4.sup.-]] = [HP[O.sub.4.sup.2-]] = 0.1 M. Potassium phosphate monobasic (K[H.sub.2] P[O.sup.4], MW = 136) and potassium phosphate dibasic ([K.sub.2] HP[O.sup.4], MW = 174) are typically used to provide the [H.sub.2] P[O.sub.4.sup.-] and HP[O.sub.4.sup.2-], respectively.

If the molar concentration of each form is 0.1 M and 1 L of buffer is required, then 0.1 mol of each chemical is used:

K[H.sub.2] P[O.sub.4] = 136 g [mol.sup.-1] x 0.1 mol = 13.6 g

[K.sub.2] HP[O.sup.4] = 174 g [mol.sup.-1] x 0.1 mol = 17.4 g

So, to make this buffer, dissolve 13.6 g of K[H.sub.2] P[O.sub.4] and 17.4 g of [K.sub.2]HP[O.sub.4] in water and dilute to 1 L.

An alternative method is to make equimolar stock solutions of the acid and base conjugates of a buffering system and then add appropriate volumes to achieve the specific pH values at the required concentration.

Example 10-14

Prepare 1 L of 0.02 M phosphate buffer, pH 6.8 from solutions of 0.2 M K[H.sub.2] P[O.sub.4] and [K.sub.2]HP[O.sub.4].

Solution

The stock concentrations are 10 times more concentrated than the final buffer concentration, so you can make 100 mL of the stock at the appropriate pH and dilute it 10-fold with little effect on pH.

pH = [pK.sub.a] + log [[A.sup.-]] / [HA]

6.8 = 7.2 + log [HP[O.sub.4.sup.2-]] / [[H.sub.2]P[O.sub.4.sup.-]]

-0.4 = log [HP[O.sub.4.sup.2-] / [[H.sub.2] P[O.sub.4.sup.-]]

The total volume of stock solutions used will be 100 mL. Simply solve the ratio for either the acid or base form on a volume basis.

mL HP[O.sub.4.sup.2-] + mL [H.sub.2] P[O.sub.4.sup.-] = 100 mL

mL HP[O.sub.4.sup.2-] = 100 mL - mL [H.sub.2] P[O.sub.4.sup.-]

-0.4 = log (100 mL - mL [H.sub.2] P[O.sub.4.sup.-]) / (mL [H.sub.2] P[O.sub.4.sup.-])

[10.sup.-0.4] = 0.398 = (100 mL - mL [H.sub.2]P[O.sub.4.sup.-]) / (mL [H.sub.2] P[O.sub.4.sup.-])

0.398 (mL [H.sub.2] P[O.sub.4.sup.-] = (100 mL - mL [H.sub.2] P[O.sub.4.sup.-])

1.398 (mL [H.sub.2] P[O.sub.4.sup.-]) = 100 mL

mL [H.sub.2] P[O.sub.4.sup.-] = 71.54 mL

mL HP[O.sub.4.sup.2-] = 28.5 mL

To make this buffer, add 28.5 mL of 0.2 M [K.sub.2] HP[O.sub.4] to 71.5 mL of 0.2 M K[H.sub.2] P[O.sub.4] and dilute to 1 L.

An advantage of using stock solutions to prepare buffers is that once you've calculated the appropriate volumes of acid and base stock for a particular pH, you should never have to calculate them again. Appendix 2 has recipes for preparing various buffers spanning the range of pH typical in soil environments using this approach.

A third alternative for buffer preparation is to use the appropriate concentration of acid or base for a buffering system and titrate with a strong acid or base (HCl or NaOH, for example) until the appropriate pH is obtained.

Example 10-15

Prepare 1 L of 0.1 M phosphate buffer, pH 7.3, starting with K[H.sub.2] P[O.sub.4] and a solution of 1 M NaOH.

Solution

The molecular weight of K[H.sub.2] P[O.sub.4] is 136 g mol. One liter of a 0.1 M buffer therefore requires:

(1 L) x (0.1 mol K[H.sub.2] P[O.sub.4] [L.sup.-1]) = 0.1 mol K[H.sub.2] P[O.sub.4]

(0.1 mol K[H.sub.2] P[O.sub.4]) x (136 g K[H.sub.2] P[O.sub.4] [mol.sup.-1]) = 13.6 g K[H.sub.2] P[O.sub.4]

All of the phosphate is supplied by K[H.sub.2] P[O.sub.4]. The pH of this solution should be approximately 4.1 since

pH = ([pK.sub.a], + PHA) / 2

So, to arrive at the appropriate pH, some of the [H.sub.2] P[O.sub.4.sup.-] must be converted to HP[O.sub.4.sup.2-] by the addition of base. How much depends on the ratio described by the Henderson-Hasselbach equation:

pH = p[K.sub.a] + log [A-] / [HA]

7.3 = 7.2 + log [HP[O.sub.4.sup.2-]] / [[H.sub.2] P[O.sub.4.sup.-]]

0.1 = log [HP[O.sub.4.sup.2-] / [[H.sub.2] P[O.sub.4.sup.-]]

[10.sup.0.1] = [HP[O.sub.4.sup.2-]] / [[H.sub.2]P[O.sub.4.sup.-]]

1.26 = [HP[O.sub.4.sup.2-]] / [[H.sub.2]P[O.sub.4.sup.-]]

To get the appropriate pH there must be 1.26 times as much HP[O.sub.4.sup.2-] in solution as [H.sub.2] P[O.sub.4.sup.-].

HP[O.sub.4.sup.2-] + [H.sub.2] P[O.sub.4.sup.-] = 0.1 M

1.26 [H.sub.2] P[O.sub.4] - = HP[O.sub.4.sup.2-]

[H.sub.2] P[O.sub.4.sup.-] + 1.26 [H.sub.2] P[O.sup.4.sup.-] = 0.1 M

2.26 [H.sub.2] P[O.sub.4.sup.-] = 0.1 M

[H.sub.2] P[O.sub.4.sup.-] = 0.044 M

HP[O.sub.4.sup.2-] = 0.056 M

(1 L) x (0.056 mol [L.sup.-1] HP[O.sub.4.sup.2-]) = 0.056 mol [H.sub.2] P[O.sub.4.sup.-] that must be converted to HP[O.sub.4.sup.2-] by titration with NaOH

0.056 mol NaOH/1 M NaOH = 0.056 L NaOH

56 mL of 1 M NaOH are required for the titration.

To make the buffer, dissolve 13.6 g of K[H.sub.2] P[O.sub.4] in a small amount of water, add 60 mL of 1 M NaOH, and dilute to 1 L with additional water.

Sample Questions

pH, Acids, and Bases

1. For the following mixtures, identify the conjugate acid and base.

A. [H.sub.2]O + HCl

B. [H.sub.2]O + [H.sup.2] P[O.sub.4.sup.-]

C. N[H.sub.3] + [H.sub.2]O

D. C[O.sub.3.sup.2-] + [H.sub.2]O

2. What is the normality (N) of the following solutions?

A. 0.5 M HCl

B. 3 x [10.sup.-5] M [H.sub.2] S[O.sub.4]

C. [10.sup.-3] M HN[O.sub.3]

D. 4 M N[H.sub.3]

E. 1 x [10.sup.-6] M [H.sub.2] C[O.sub.3]

F. 1 x [10.sup.-4] M [H.sub.3] P[O.sub.4]

G. 1 x [10.sup.-2] M [Ca.sup.2+]

H. 4 x [10.sup.-3] M [Al.sup.3+]

I. 1 x [10.sup.-7] M [Na.sup.+]

3. What is the pH of the following solutions?

A. 5 M HCl

B. 2 x [10.sup.-5] M [H.sub.2] S[O.sub.4]

C. 4 M HN[O.sub.3]

D. 1 x [10.sup.-5] M HCl

E. 2 X [10.sup.-5] M NaOH

4. What is the [H.sup.+] and O[H.sup.-] concentration in solutions with the following pH values?

A. 3.7

B. 6.4

C. 7.8

D. 8.7

E. 11.4

5. What is the final pH of a mixture of 100 mL of 0.3 M NaOH and 200 mL of 0.2 M [H.sub.2] S[O.sub.4]?

6. If you add 0.1 g of NaOH to 1.0 L of distilled water at pH 7, what will the final pH be?

7. If concentrated HCl has a density of 1.19 g [cm.sup.-3] (mL) and contains 37.5 percent HCl by weight, how many moles of HCl are in 25 mL of concentrated HCl and what would be the pH if 25 mL were diluted to 300 mL with water at pH 7.0?

8. What is the pH and pOH of a 0.003 M solution of HN[O.sub.3], a strong acid?

9. What is the concentration of HCl in an HCl-amended solution that has a pH of 3.4?

10. What is the [[H.sup.+]] in a 3 X [10.sup.-4] N solution of [H.sub.2] S[O.sub.4]?

Neutralization

1. How much 0.1 N HCl would be required to titrate 50 mL of 0.25 N NaOH to neutrality? What size of beaker would you need for your titration?

2. How much 0.2 M [H.sub.2] S[O.sub.4] is required to neutralize 50 percent of the O[H.sup.-] present in 600 mL of 0.2 M NaOH?

3. What is the pH of a solution in which 0.4 M [H.sub.2] S[O.sub.4] is used to completely neutralize 600 mL of 0.3 M KOH?

Weak Acids

1. What is the [pK.sub.a] of a 0.3 M weak acid solution (HA) that is 5 percent dissociated?

2. How many milliliters of 0.2 M KOH are required to titrate 300 mL of 0.4 M acetic acid to neutrality?

3. What is the final pH of a solution obtained by mixing 150 mL of 0.3 M NaOH with 400 mL of 0.2 M acetic acid, assuming the p[K.sub.a] of acetic acid is 4.77?

4. If the pH of a borate buffer is 9.2, how much [H.sub.2] B[O.sub.3.sup.-] and HB[O.sub.3.sup.2-] are in a 0.3 M buffer that has a pH of 9.5?

Buffers and Buffering

1. How do you prepare 5 L of 0.2 M glycine buffer at pH 10 if your starting ingredients are glycine (75 g [mol.sup.-1]) and 2 M KOH?

2. What is the relative proportion of mono-and dibasic citrate in a 0.3 M citrate buffer at pH 3.76, 4.76, and 5.76?

3. Prepare 2 L of 0.03 M phosphate buffer, pH 3.0, from stock solutions of 0.5 M [H.sub.3] P[O.sub.4] and K[H.sub.2] P[O.sub.4].

4. Prepare 50 mL of 0.05 M phosphate buffer, pH 11, starting with [K.sub.2] HP[O.sub.4] and a solution of 0.3 M KOH.

Reference

Segal, 1. H. (1976). Biochemical calculations (2nd ed., p. 441). New York: John Wiley & Sons.
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Title Annotation:Section III: Problem Solving in Soil Biochemistry
Publication:Math for Soil Scientists
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Date:Jan 1, 2006
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