Titration Calculations A volume of 35.27 mL of 0.10 M sodium hydroxide is requir

Question

Titration Calculations A volume of 35.27 mL of 0.10 M sodium hydroxide is required to titrate 200.0 mL of a solution containing an unknown quantity of citric acid, H3CGH,o (a) Write the balanced equation for the COMPLETE neutralization of citric acid (shown below) with NaOH (b) What was the original molarity of the citric acid analyte? (c) How many grams of citric acid were present in the flask before the titration? You are titrating 40.0 mk of 0.100 M hydrochloric acid with 0.100 M sodium hydroxide. (a) Identify the titrant and the analyte for this titration. (b) Write the balanced equation for the neutralization reaction that occurs as you add titrant. (c) What volume of 0.100 M NaOH is required to reach the equivalence point? (d) Determine the overall volume and pH of the resulting solution (in the flask) after the following volumes of titrant have been added: 0.00 mL, 10.00 mL, 40.00 mL, 50.00 mL

Explanation / Answer

ANSWER

Q1.

(a) The complete balanced equation for this neutralisation equation is given below

H3C6H5O7 + 3NaOH --------> C6H5O73- (aq) + 3Na+(aq) + 3H2O(l)

(b) Molarity of citric acid = No. of moles citric acid / volume in liters

Moles of citric acid = moles of NOH consumed

Moles of NaOH consumed = Molarity X Volume in Liters = 0.10 X 0.03527 = 0.003527 moles

35.27mL = 0.03527L

Thus molarity of citric acid = 0.003527 / 0.2L = 0.18M

(c) No. of grams of citric acid = No. of moles X Molar mass = 0.003527 X 192.124 = 0.68g

Q2.

(a) HCl is the analyte and NaOH titrant.

(b) Balanced equation is :

HCl + NaOH -------> NaCl + H2O

(c) The required volume of NaOH can be had from the following equation:

MV (NaOH) = MV(HCl)

M and V = molarity and volume respectively.

V(NaOH) = MV(HCl) / M (NaOH) = 0.10 X 40mL / 0.10 = 40mL

(d) When 0 mL of NaOH is added there will be only 0.10M HCl

pH = - log[H+] = -log(0.1) = 1

when 10mL of NaOH is added:

Moles of NaOH added = ) = 0.1 X 10mL = 1.0mmol

1.0mmol of HCl will be netralised

Moles of HCl remained = Original moles - moles reacted = (0.1 X 40mL) - 0.1 = 3.9mmoles

Total volume = 40mL + 10mL = 50mL

Molarity of Remained HCl Solution = 3.9 mmol / 50mL = 0.078M

pH = -log(0.078) = 1.1

When 40mL of NaOH is added

40mL of 01M NaOH will correspond to equivalence point. Thust there will be no HCl , Solution will be neutralsed.

Hence pH will be 7.

When 50mL of NaOH is added:

Moles of NaoH added = 0.1 X 50mL = 5.0mmol

Out of 5mmol of NaOH 4.0mmol will react with HCl and 1.0mmol of NaOH will remain unreacted in the solution.

New volume = 40mL + 50mL = 90mL

Molarity of NaOH = 1.0mmol / 90mL = 0.01M

NaOH will furnish OH-

pOH = -log(0.01) = 2

pH = 14 - 2 = 12

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