Process, Observations, and Reactions
1. The indicator used, i.e., phenolphthalein has an endpoint from colorless to pink. This means that it is colorless in acids and turns pink in bases. Thus, as it is added in the flask at the start of the procedure, the acidic solution remains colorless. When a specific volume of NaOH has reacted with the acid, the color change is observed; it is very crucial to stop the addition of the titrant at this point, as any extra amount of the base can give a wrong reading, thus, leading to entirely wrong calculations. At this point, the reaction being complete, the indicator detects the presence of adequate amount of a base solution in the mixture and hence, turns pink.
2. A reaction between sulfuric acid and sodium hydroxide is of an acid-base type, or is also known as a neutralization reaction. In this process, both compounds undergo a reaction to neutralize the acid and base properties. The products of this process are salt and water. The former is produced when the cation (a positively charged ion) of the base combines with the anion (a negatively charged ion) of the acid. In the language of chemistry, the cation of the base NaOH, i.e., Na+, combines with the anion of the acid H2SO4, i.e., SO42-. This gives out Na2SO4. Water is formed as a result of the reaction between the cation of the acid (H+) that combines with an anion of the base (OH-).
The balanced equation of this reaction is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O ---- (equation 1)
3. In this process, 2 moles (the molecular weight of a substance expressed in grams) of sodium hydroxide (NaOH) combine with one mole of sulfuric acid (H2SO4). This results in the formation of two moles of water (H2O) and one mole of sodium sulfate (Na2SO4). The net ionic equation (a chemical equation in which electrolytes are written as dissociated ions) can be explained as follows:
The complete balanced equation for the reaction between sulfuric acid and sodium hydroxide is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Thus, writing this in a full ionic reaction form , we get:
2H+ + SO42- + 2 Na+ + 2OH- → 2Na+ + SO42- + 2H2O
So, the net ionic equation of sulfuric acid and sodium hydroxide, after crossing out the spectator ions, is:
2H+ + 2OH- → 2H2O
2. A reaction between sulfuric acid and sodium hydroxide is of an acid-base type, or is also known as a neutralization reaction. In this process, both compounds undergo a reaction to neutralize the acid and base properties. The products of this process are salt and water. The former is produced when the cation (a positively charged ion) of the base combines with the anion (a negatively charged ion) of the acid. In the language of chemistry, the cation of the base NaOH, i.e., Na+, combines with the anion of the acid H2SO4, i.e., SO42-. This gives out Na2SO4. Water is formed as a result of the reaction between the cation of the acid (H+) that combines with an anion of the base (OH-).
The balanced equation of this reaction is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O ---- (equation 1)
3. In this process, 2 moles (the molecular weight of a substance expressed in grams) of sodium hydroxide (NaOH) combine with one mole of sulfuric acid (H2SO4). This results in the formation of two moles of water (H2O) and one mole of sodium sulfate (Na2SO4). The net ionic equation (a chemical equation in which electrolytes are written as dissociated ions) can be explained as follows:
The complete balanced equation for the reaction between sulfuric acid and sodium hydroxide is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Thus, writing this in a full ionic reaction form , we get:
2H+ + SO42- + 2 Na+ + 2OH- → 2Na+ + SO42- + 2H2O
So, the net ionic equation of sulfuric acid and sodium hydroxide, after crossing out the spectator ions, is:
2H+ + 2OH- → 2H2O
Calculations and Inferences
Let's find out the molarity or concentration of sulfuric acid solution from the given and observed data (molarity and observed volume of sodium hydroxide).
For example, during three rounds of the experiment, the amount of NaOH needed to react with sulfuric acid is 12 ml, 13 ml, and 12.5 ml, respectively. Thus, by taking their average, 12.5 ml of NaOH neutralized the acid with the unknown concentration. Suppose the molarity of sodium hydroxide is 0.1 mol/L. The formula for calculating the number of moles of a solution is:
Number of Moles (N) = Volume (V) x Molarity (M)
Thus, the formula becomes,
N = 0.0125 x 0.1 = 0.00125 moles
Now, according to the equation 1 mentioned above, exactly half the number of moles of sulfuric acid take part in the reaction; i.e., the number of moles required for the neutralization process are:
N (for sulfuric acid) = 0.00125/2 = 0.000625 moles
The volume of sulfuric acid used in this experiment is 10 ml. By rearranging the above formula, the molarity or concentration of H2SO4 is calculated:
Molarity (M) = Number of Moles (N)/Volume (V) M = 0.000625/0.01 = 0.0625mol/L
Thus, the molarity or concentration of sulfuric acid in the above-described experiment is 0.0625 mol/L.
For example, during three rounds of the experiment, the amount of NaOH needed to react with sulfuric acid is 12 ml, 13 ml, and 12.5 ml, respectively. Thus, by taking their average, 12.5 ml of NaOH neutralized the acid with the unknown concentration. Suppose the molarity of sodium hydroxide is 0.1 mol/L. The formula for calculating the number of moles of a solution is:
Number of Moles (N) = Volume (V) x Molarity (M)
Thus, the formula becomes,
N = 0.0125 x 0.1 = 0.00125 moles
Now, according to the equation 1 mentioned above, exactly half the number of moles of sulfuric acid take part in the reaction; i.e., the number of moles required for the neutralization process are:
N (for sulfuric acid) = 0.00125/2 = 0.000625 moles
The volume of sulfuric acid used in this experiment is 10 ml. By rearranging the above formula, the molarity or concentration of H2SO4 is calculated:
Molarity (M) = Number of Moles (N)/Volume (V) M = 0.000625/0.01 = 0.0625mol/L
Thus, the molarity or concentration of sulfuric acid in the above-described experiment is 0.0625 mol/L.
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