The titration of a 25.0mL sample of 0.100 M acetic acid using 0.100 M NaOH as th

Question

The titration of a 25.0mL sample of 0.100 M acetic acid using 0.100 M NaOH as the titrant. The value of the equilibrium constant for HC2H3O is Ka=1.8x 10-5. Calculate the pH of the acid solution before any NaOH has been added. Calculate the pH of the analyte after 10.00mL of 0.100 M NaOH has been added. Calculate the pH of the analyte of the equivalence point? Calculate the pH of the analyte after adding 26.00 mL of 0.100 M NaOh? Please show all work.

Explanation / Answer

Moles acetic acid = 0.0250 L x 0.100 M = 0.00250 moles NaOH = 0.006 L x 0.100 M = 0.0006 CH3COOH + OH->> CH3COO - + H2O moles CH3COOH = 0.00250 -0.0006 = 0.0019 moles CH3COO- = 0.0006 total volume = 0.0310 L [CH3COOH]= 0.0019/ 0.0310= 0.0613 M [CH3COO-] = 0.0006/ 0.0310= 0.0184 M pKa = 4.7 pH = pKa + log [CH3COO-] / [CH3COOH] pH = 4.7 + log 0.0184/ 0.0613 = 4.2 Answer

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