Question

Lactic acid, \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), produced in \(1 \mathrm{~g}\) sample of muscle tissue was titrated using phenolphthalein as indicator against \(\mathrm{OH}^{-}\) ions which were obtained by the electrolysis of water. As soon as \(\mathrm{OH}^{-}\) ions are produced, they react with lactic acid and at complete neutralization, immediately a pink colour is noticed. If electrolysis was made for 1158 s using \(\quad 6\) \(50.0 \mathrm{~mA}\) current to reach the end point, what was the percentage of lactic acid in muscle tissue? (a) \(5.4 \%\) (b) \(2.7 \%\) (c) \(10.8 \%\) (d) \(0.054 \%\)

Short answer

The percentage of lactic acid in muscle tissue is 2.7%.

Step-by-step solution

Calculate the amount of charge passed

First, calculate the total charge (Q) passed during the electrolysis using the formula Q = It, where I is the current in amperes and t is the time in seconds. Here, I = 50.0 mA (which is 0.0500 A because 1 A = 1000 mA) and t = 1158 s.

Determine the moles of electrons

Use the charge calculated in Step 1 and Faraday's constant (\( F = 96485 \text{ C/mol} \text{ of electrons} \) ) to find the moles of electrons that have been transferred. The formula is \( n = \frac{Q}{F} \) where n is the number of moles of electrons.

Determine the moles of lactic acid neutralized

Each mole of lactic acid reacts with 1 mole of \(\mathrm{OH}^-\) ions (1:1 molar ratio). The number of moles of \(\mathrm{OH}^-\) produced is equal to the moles of electrons since each electron produces one \(\mathrm{OH}^-\) ion. Therefore, the moles of lactic acid neutralized equals the moles of electrons.

Calculate the mass of lactic acid

Using the molar mass of lactic acid (\(\mathrm{M_{HC_3H_5O_3}} = 90.08 \text{ g/mol}\) ), you can find the mass of lactic acid by multiplying the moles of lactic acid by its molar mass using the formula mass = moles × molar mass.

Calculate the percentage of lactic acid in the tissue

Finally, calculate the percentage of lactic acid in the 1 g sample of muscle tissue by taking the mass of lactic acid from Step 4 and dividing it by the sample mass (1 g), then multiply by 100 to get the percentage.

Key concepts

Electrolysis Calculations

Electrolysis is a chemical process that uses electricity to drive a non-spontaneous reaction. It's particularly important in the field of analytical chemistry when dealing with reactions like the neutralization of lactic acid with hydroxide ions, which are produced through the electrolysis of water.

To calculate the different aspects of an electrolysis reaction, a good understanding of the principles underlying electrolysis is essential. In a titration problem involving electrolysis, the initial step requires determining the total electric charge passed through the electrolyte. This is found using the formula \(Q = It\), where \(Q\) is the charge in coulombs (C), \(I\) is the current in amperes (A), and \(t\) is the time in seconds (s).

Once you have the total charge, you can further delve into the stoichiometry of the reaction to see how much of a substance is produced or consumed during the electrolysis. The calculation is intrinsically related to Faraday's laws, molar mass, and stoichiometry - all of which combine to give us a holistic view of what's happening at the molecular level during the reaction.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis provide the quantitative relationship between the amount of electric charge used during electrolysis and the amount of substance that is oxidized or reduced at the electrodes.

The first law states that the chemical amount of a substance altered at an electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte. This can be formulated as \(m = (Q/F) \times M\), where \(m\) is the mass of the substance altered, \(Q\) is the total charge passed, \(F\) is Faraday’s constant (approximately 96485 C/mol), and \(M\) is the molar mass of the substance.

The second law of Faraday states that, when the same amount of electric charge is passed through multiple electrolytes, the amount of substance altered is directly proportional to its equivalent weight. This law is particularly useful when dealing with electrolytes in a series.

Molar Mass Calculation

The molar mass of a compound is a critical piece of information in stoichiometry and electrolysis calculations. It is the mass of one mole of that substance, typically expressed in grams per mole (g/mol).

For a substance such as lactic acid, with the chemical formula \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), you would calculate its molar mass by summing the atomic masses of each constituent atom, derived from the periodic table. For example, molar mass of lactic acid would consider the weights of carbon, hydrogen, and oxygen in the given quantities in the molecular formula.

Knowing the molar mass allows us to convert between the mass of a substance and the amount of substance in moles, paving the way for quantity comparisons and further stoichiometric calculations in a titration problem.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the calculation of the quantities of reactants and products involved in chemical reactions. It's fundamental to titration problems like the one involving lactic acid and the produced hydroxide ions from electrolysis.

The stoichiometry of a reaction outlines the precise reactant-to-product ratios. In the provided exercise, for instance, the stoichiometry of lactic acid reacting with hydroxide ions is a 1:1 molar ratio. This means that one mole of lactic acid reacts with one mole of hydroxide ions. Using this ratio, we can calculate the amount of lactic acid neutralized by hydroxide ions produced during electrolysis.

Armed with the molar mass and the mole ratio, the final step is to convert moles back into grams to find out the actual mass of the lactic acid consumed, and consequently, its percentage in the given muscle tissue sample. The ability to navigate through these quantitative aspects is vital for accurately solving and understanding the nuances of titration problems in chemistry.