You have 425 mL of an 0.35 M acetic acid solution. What volume (V) of 1.10 M NaO

Question

You have 425 mL of an 0.35 M acetic acid solution. What volume (V) of 1.10 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.34? (The pKa of acetic acid is 4.76.)

Explanation / Answer

The reaction for this is: CH3COOH + OH(-) ----> CH3COO(-) + H2O Initial mol of CH3COOH: 0.425 L x 0.35 M = 0.15 mol In the reaction, NaOH will be the limiting reagent. x mols of OH(-) will be used up, meaning that (0.15 - x) mol of CH3COOH will remain and x mol of CH3COO(-) will be created. The final concentration of CH3COOH is (0.15 - x) mol/(0.425 L + x/1.10) The final concentration of CH3COO- is x/(0.425 L + x/1.10). x/1.10 M represents the volume of NaOH added to the solution. Set up Henderson Hasselbalch equation given these things: pH = pKa + log([CH3COO-]/[CH3COOH]) 5.34 = 4.76 + log(x/(0.15-x)) : Note that the denominator of (0.425 L + x/1.10 M) will cancel out in the fraction Solve for x to get x = 0.12 mol. This means 0.12/1.1 = 0.11 L or 110 mL of 1.10 M NaOH is added.

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