Lactic acid (CH3CHOHCOOH) is found in sour milk and in muscles after activity. (Ka for lactic acid = 1.4 × 10-4.)If 2.25 g of CH3CHOHCOONa, sodium lactate, is added to 500 mL of 0.130 M lactic acid,What is the pH of the resulting buffer solution?
Question
Lactic acid (CH3CHOHCOOH) is found in sour milk and in muscles after activity. (Ka for lactic acid = 1.4 × 10-4.)If 2.25 g of CH3CHOHCOONa, sodium lactate, is added to 500 mL of 0.130 M lactic acid,What is the pH of the resulting buffer solution?
Solution
To solve this problem, we will use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). Here, [A-] is the concentration of the base (sodium lactate) and [HA] is the concentration of the acid (lactic acid).
Step 1: Calculate the pKa from the Ka pKa = -log(Ka) = -log(1.4 × 10^-4) = 3.85
Step 2: Calculate the concentration of sodium lactate ([A-]) First, convert the mass of sodium lactate to moles: 2.25 g / (CH3CHOHCOONa molar mass 112.06 g/mol) = 0.020 mol Then, convert the volume of the solution to liters: 500 mL = 0.5 L So, [A-] = 0.020 mol / 0.5 L = 0.040 M
Step 3: The concentration of lactic acid ([HA]) is given as 0.130 M.
Step 4: Substitute these values into the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) = 3.85 + log(0.040/0.130) = 3.85 - 0.48 = 3.37
So, the pH of the resulting buffer solution is 3.37.
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