A student plans to titrate 25.00 mL of a 0.184 M HCOOH solution with 0.141 M NaO
ID: 731757 • Letter: A
Question
A student plans to titrate 25.00 mL of a 0.184 M HCOOH solution with 0.141 M NaOH solution.a) Calculate the volume of NaOH (in mL) that will be needed to reach the equivalence point.
b) Calculate the anticipated pH at the equivalence point.
*Please show all work and calculations.
Explanation / Answer
42.2 X 0.052 = Vol. NaOH X 0.0372 Vol.NaOH = 2.1944/0.0372 = 58.99 ml so volume of NaOH recquired to reach equivalence point = 58.99 ml number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944 now CH3COOH and NaOH reacts to give CH3COONa according to the reaction : CH3COOH + NaOH ------> CH3COONa + H2O 1 mole of CH3COOH reacts with 1 mole of NaOH to give 1 mole of CH3COONa so 2.1944 millimoles of CH3COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH3COONa... so all the acid (CH3COOH) and base (NaOH) has been converted into salt (CH3COONa) so there is no acid or base left ...... now molarity of CH3COONa = number of millimoles of CH3COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M and as CH3COONa is a salt of weak acid (CH3COOH) and strong base (NaOH) ... so using the hydrolysis equation : pH = 1/2 [ pKw + pKa + log c ] Ka for acetic acid = 1.75 X 10^-5 so pKa = -log (1.75 X 10^-5) = 4.74 Kw = 10^-14 so pKw = -log 10^-14 = 14 c = 0.02169 so log c = log 0.02169 = -1.66 putting the values.... pH = 1/2 [14 + 4.74 - 1.66 ] pH = 1/2 [ 17.08] = 8.54
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