In Part A of Experiment 3, a solution containing 1.5 g of CH_3COONa. 3H_2O cryst

Question

In Part A of Experiment 3, a solution containing 1.5 g of CH_3COONa. 3H_2O crystals (MW = 136.08) was added to 2.50 mL of 0.40 M HCl, resulting in the formation of acetic acid (CH_3COOH). Assuming the reaction proceeded to completion, calculate the number of moles of acetic acid and acetate ion in solution. Enter your answers with the correct number of significant figures in the appropriate boxes, and in the space below. Calculate the theoretical pH of the resulting solution., and enter your final answer in the box provided. Briefly explain why the actual pH of the resulting solution may have differed from the theoretical pH by more than Plusminus 0.02 (the error associated with the pH meter).

Explanation / Answer

mol of conjugate = mass/MW = 1.5/136.08 = 0.011022 mol

mol of HCL = MV = (2.*10^-3)(0.40) =0.0008 mol of

ratio is 1:1 so...

mol of conjugte left = 0.011022 -0.0008 = 0.010222

mol of acid formed = 0+0.0008 = 0.0008

mol of CH3COOH formed --> 0.0008 mol

mol of CH3COO- left --> 0.010222 mol

b)

pH theoretical

this will be a buffer

so apply buffer equation Henderson Haselbach

pH = pKa + log(A-/HA)

pKa for acid = 4.75

pH = 4.75 - log(CH3COO-/CH3COOH )

pH = 4.75 - log(0.010222 /0.0008 )

pH = 3.64

c)

the pH may differ since the solution has water increase volume, i.e. also activity of ions is present

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