question archive The NaOH solution used to determine titratable acidity was found to have a concentration of 0
The NaOH solution used to determine titratable acidity was found to have a concentration of 0
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The NaOH solution used to determine titratable acidity was found to have a concentration of 0.096N upon standardization with KHP. Calculate the titratable acidity (in tartaric acid equivalents) for a 10 mL wine sample if 5.6 mL of this NaOH solution is used.
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- Titratable acidity is a total amount of acid in the solution as determined by the titration using a standard solution of sodium hydroxide (titrant). Molar mass (C4H6O6)=4 x M(C)+6 x M(H)+6 x M(O)=4 x 12+6 x 1+6 x 16=150 g/mole.
- Multiply the volume of the standard solution of NaOH by its concentration to determine the number of moles of the titrant used for the titration. Number of moles=0.0096 L x 0.1 mole/L=0.00096 moles
- In our example, the volume of the NaOH solution used is 5.6 ml or 0.0056 L. Hence, number of moles (NaOH)=Number of moles=Volume (in L) x molar (mole/L) concentration.
- Write down the chemical reaction the titration is based on. In our example, it is a neutralization reaction expressed as C4H6O6+2NaOH= C4H4O6Na2+2H2O.
- Determine the number of moles of the acid using the equation in Step 3. In our example, according to that equation, one molecule of the acid reacts with two molecules of NaOH. Thus, 0.0096 moles of NaOH (Step 2) would interact with 0.0048 moles of tartaric acid.
- Divide the number of moles of the acid (Step 4) by the aliquot volume and then multiply by 100 to calculate the acid amount in 100 ml of the solution. In our example, amount (C4H6O6)=0.00048moles x 100 ml/10 ml=0.0048 moles.
- Multiply the acid amount in 100 ml (Step 5) by its molar mass (Step 1) to calculate titratable acidity (in g/100 ml). In our example, titratable acidity=0.0048 x 150=0.72 g/100 ml.
