The Lester (Ratio computations and preparations of statements) Lester Fredrick c

Question

The Lester (Ratio computations and preparations of statements) Lester Fredrick corporation has in recent years maintained the following relationship among the data on its financial statement :
1. Gross profit rate on net sale 30%
2. Net profit margin on net sales 8%
3. Rate of selling expenses to net sales 15%
4. Accounts receivable turnover 8per year
5. Inventory turnover 10per year
6. Acid-test ratio 2 to 1
7. Current ratio 3 to 1
8. Quick asset composition:8% cash,27% marketable securities, 65% accounts receivable
9. Asset turnover 2 per year
10. Ratio of total assets to intangible assets 20 to 1
11. Ratio of accumulated depreciation to cost of fixed assets 1 to 3
12. Ratio of accounts receivable to accounts payable 2 to 1
13. Ratio of working capital to stockholders equity 1 to 1.95
14. Ratio of total debt to stockholders equity 1 to 3
The corporation had a net income after tax of $520000 for the year ended December 31,2003,which resulted in earnings of 9.74 per share of common stock. Addition information includes the following:
1. Capital stock authorized, issued(all in 1982), and outstanding:
Common, $10 per share par value, issued at 10% premium
Preferred, 11% nonparticipating,$100 per share par value, issued at a 10% premium
2. Market value per share of common at December 31,2008:$119.52
3. Preferred dividends paid in 2003: $33000
4. Times interest earned in 2003: 28.73
5. The amounts of the following were the same at December 31,2008,as at January 1,2003: inventory, account receivable, 10% bonds payable due 2006,and total stockholders

Explanation / Answer

It will be noted that many buffer Problems involve a determination of the ratio of the [salt] to [acid], each of which must be expressed in the same units, and the total of the [salt] and [acid] together. The general situation calls, therefore, for two simultaneous equations

[salt]/[acid]=a, [salt]+[acid]=b

It is essential in any buffer problem to identify from the given description the values of pKa, pH, a, and b, which can be expressed in many ways. pKa can be given as Ka, or the equilibrium concentrations of the weak acid. pH can be given as [H+], pOH, or [OH-]. [Salt] or [acid] may be derived by titration, but in all cases the four unknowns must be obtained from the given information or by calculation. The most comprehensive buffer problem asks for the preparation of a volume of buffer of a certain concentration and pH from a weak acid of given pKa, and a strong base. The general approach may be by the following steps.

1. Since pH and pKa are given, determine the [salt]/[acid] ratio, which is a pure number.

The concentration of the buffer is usually expressed as the sum of the salt and acid forms.

If the salt (or the acid) concentration is given, then the corresponding acid (or salt) can be calculated directly from the ratio of [salt] to [acid] in step 1. If the total concentration, [salt] + [acid], is given, then the solution of the two simultaneous equations is required. Note that the unit of concentration of the salt and acid so calculated will be the one used for the [salt] + [acid] concentration. For example, if

[salt]/[acid]=2

[salt] + [acid]=0.3 M

Then, since [salt] = 2[acid]

and by substitution ,

2[acid] + [acid] = 0.3 M

[acid] = 0.3/3 M = 0.1 M

Now, if we substitute this value ,

[salt] = 2 x 0.1 M = 0.2 M

Check these values in (i) and (ii):

[salt[/[acid]=0.2 M/0.1 M = 2

[salt] + [acid] = 0.2 M + 0. 1 M = 0.3 M

3. From the concentrations of salt and acid the actual amounts of salt and acid can be calculated for the volume required from their molecular weights. In the most general case, where the given materials are pure acid and pure base, the salt will be generated from equivalent amounts of acid and base; it is necessary to measure an amount of pure acid equivalent to salt and acid and add to it an amount of base equivalent to the salt.

In the example given in step 2 above, if 1 liter of buffer were required and if the acid were monobasic, then to 0.3 mole of acid would be added 0.2 mole of base and the volume brought to 1 liter. This would give 0.2 M salt and leave 0. 1 M free acid.

An alternative approach to that developed above for the solution of problems involving equilibria is to use the fraction of dissociation, a, which is defined for any system as

a = (moles of HA dissociated)/( total moles of species containing A).

= (moles HA dissociated)/(AT).

For the dissociation of a weak acid

[At] = [HA] + [A-],

then

a = [A-]/[HA] + [A-],

This applies equally to the salt of a weak acid and, in such cases as buffer systems,

[salt]/[acid] = [A-]/[HA] = a/1

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