Question

the diprotic acid lactic acid is produced as a result of anaerobic respiration. raquel titrates the lactic acid in a yogurt sample according to the equation below. \\(\ce{h2c3h4o3 + 2koh \
ightarrow k2c3h4o3 + 2h2o}\\) \\(16.35\space ml\\) of \\(0.05\space m\\) \\(\ce{koh}\\) reacts with \\(40.00\space ml\\) of the analyte sample. what is the concentration of the lactic acid in the sample? \\(\ce{? \times 10^{?} m h2c3h4o3}\\)

Explanation:

Step1: Recall the formula for titration (moles of acid and base)

In a titration, the moles of acid and base are related by the stoichiometry of the reaction. The formula is \( n_1M_1V_1 = n_2M_2V_2 \), where \( n \) is the stoichiometric coefficient, \( M \) is molarity, and \( V \) is volume (in liters). For the reaction \( \ce{H2C3H4O3 + 2KOH -> K2C3H4O3 + 2H2O} \), the stoichiometric coefficient of lactic acid (\( n_1 \)) is 1, and for KOH (\( n_2 \)) is 2.

Step2: Convert volumes to liters

\( V_1 \) (lactic acid) = \( 40.00 \, \text{mL} = 40.00 \times 10^{-3} \, \text{L} \)
\( V_2 \) (KOH) = \( 16.35 \, \text{mL} = 16.35 \times 10^{-3} \, \text{L} \)
\( M_2 \) (KOH) = \( 0.05 \, \text{M} \)

Step3: Rearrange the formula to solve for \( M_1 \) (molarity of lactic acid)

From \( n_1M_1V_1 = n_2M_2V_2 \), we get \( M_1=\frac{n_2M_2V_2}{n_1V_1} \)

Step4: Substitute the values

\( n_1 = 1 \), \( n_2 = 2 \), \( M_2 = 0.05 \, \text{M} \), \( V_2 = 16.35 \times 10^{-3} \, \text{L} \), \( V_1 = 40.00 \times 10^{-3} \, \text{L} \)
\( M_1=\frac{2\times0.05\times16.35\times 10^{-3}}{1\times40.00\times 10^{-3}} \)

Step5: Calculate the numerator and denominator

Numerator: \( 2\times0.05\times16.35\times 10^{-3}=2\times0.05\times0.01635 = 0.001635 \)
Denominator: \( 1\times40.00\times 10^{-3}=0.04 \)
\( M_1=\frac{0.001635}{0.04}=0.0204375 \, \text{M} \)
\( 0.0204375 = 2.04375\times 10^{-2} \, \text{M} \) (rounding appropriately, we can write it as \( 2.04\times 10^{-2} \) or more accurately \( 2.04375\times 10^{-2} \))

Answer:

\( 2.04 \times 10^{-2} \) (or more precisely \( 2.04375\times 10^{-2} \))