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Acid - Base Neutralization


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Aicd-Base Neutralization

(a) 12.5 ml of a 50 ml solution of sulphuric acid containing 0.490 g of sulphuric acid completely neutralized 20 ml of a sodium hydroxide solution during titration. Determine the concentration of the sodium hydroxide solution.

(b) The hydrochloric acid (HCl) concentration in the gastric juice of a patient with duodenal ulcer is 80 x 10-3 M. The patient produces 3 liters of gastric juice per day and his doctor has prescribed a medication containing 2.6 g Al(OH)3 per 100 ml of solution, for the relief of excess stomach acidity (Al(OH)3 and Mg(OH)2 are common ingredients in medications designed to neutralize stomach acid). Determine the patient's daily dose of the prescribed medication that will neutralize the acid.

The strings:

S7P5A52 (Change - Chemical Change).

The math:
Pj Problem of Interest is of type change (chemical change).

Acid-Base Neutralization



The chemical reaction of an acid with a base is called a neutralization reaction. Salt and water are the products of a neutralization reaction. For example, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) in aqueous solution according to the following chemical equation:
HCl(aq) + NaOH(aq) -------> H2O(l) + NaCl(aq)-------(1)
In reaction (1), the acidity of hydochloric acid is neutralizaed.

Titration is a laboratory procedure for determining the amount of one substance by determining the volume of a solution of known concentration of another substance that reacts completely with the first substance. Titration is commonly used in neutralization reactions to determine the amount of base that is chemically equivalent to a given amount of acid.

An equivalent is the molecular weight or mass of an acid or base that furnished one mole of protons (H+ or one mole of hydroxl (OH- ions. For example, the equivalents contained in sulphuric acid (H2SO4) is calculated as follows:
(molecular weight of H2SO4)/(moles of proton furnished)
= 98/2 = 49. Since each mole of H2SO4 produces two protons.

(a) At complete neutralization, the number of equivalents of acid in the 12.5 ml volume is equal to the number of equivalents of base in the 20 ml volume.
normality, N = equivalents/liters.
So,NacidVacid = NbaseVbase -------- (2)

Nacid in 50 ml (0.050 l) = number of equivalents in 0.050 l/0.050
= (mass of acid/gram equivalent weight)/0.050
= (0.490/49)/0.050 = 0.200 N.

So from equation (2), Nbase = (NacidVacid)/Vbase
So, Nbase = (0.200 x 12.5)/20 = 0.125 N.
Since molarity = number of mole/liters and there is 1 ionizable OH- in NaOH:
Normality = Molarity
So, concentration of the sodium hydroxide solution is 0.125 N or 0.125 M.

(b) The neutralization reaction between HCl and Al(OH)3 is:
Al(OH)3 + 3HCl --------> AlCl3 + 3H2O -------(3)
At complete neutralization, the number of equivalents of Al(OH)3 equals the number of equivalents of HCl.
So, NHClVHCl = NAl(OH)3VAl(OH)3 -------- (4)
Where N is normality and V is volume.
Molarity of Al(OH)3 = (2.6/78)/0.100 = 0.33 M
Al(OH)3 contains 3 ionizable hydroxl (OH-) ions
So, NAl(OH)3 = 3 x 0.33 = 0.99 N
Since HCl contains 1 ionizable ion, molarity of HCl = NHCl
So, NHCl = 80 x 10-3 N
So, VAl(OH)3 = (NHClVHCl)/NAl(OH)3
= (80 x 10-3 N x 3 liters)/0.99 N = 0.240 liters = 240 ml
So, daily dose of medication is 240 ml.

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