. Find the volume of 0.1 M NaOH needed to neutralize 10 mL of 0.1M HO(CO)CH=CH(C
ID: 1027207 • Letter: #
Question
. Find the volume of 0.1 M NaOH needed to neutralize 10 mL of 0.1M HO(CO)CH=CH(CO)OH.
..in this lab you are going to make approximately 100 mL of an 0.1M acetic acid-sodium acetate buffer system that's at pH= 4.9 . you will start by making 100 mL of a 0.1 M acetic acid solution from 2.00M Acetic Acid starting solution. you will then monitor the pH of the solution and add solid sodium acetate until the pH reaches 4.9 . answer the follwing questions.
1. calculate the amount of sodium acetate in grams that you will need to add to adjust the pH to 4.9
2. calculate the volume of 0.1M NaOH that 40mL of the buffer could neutralize?
3. Calculate the volume of 0.1M HCl that 40mL of the buffer could neutralize?
Explanation / Answer
1) The given acid is a diacid, i.e, contains two –COOH groups. The acid base reaction can be shown as
HOOCCH=CHCOOH + 2 NaOH ---------> NaOOCCH=CHCOONa + 2 H2O
As per the stoichiometry of the reaction,
1 mole HOOCCH=CHCOOH = 2 moles NaOH
Mole(s) of HOOCCH=CHCOOH taken = (10 mL)*(1 L/1000 mL)*(0.1 M) = 0.001 mole.
Mole(s) of NaOH required for neutralization = 2*0.001 mole = 0.002 mole.
Volume of 0.1 M NaOH required for complete neutralization = (0.002 mole)/(0.1 M) = 0.02 L = (0.02 L)*(1000 mL/1 L) = 20 mL (ans).
2i) The pKa of acetic acid, CH3COOH is 4.75.
We wish to prepare 100 mL of the buffer where the concentration of CH3COOH is 0.1 M. The supplied stock solution is 2.00 M.
Use the Henderson-Hasslebach equation to find out the mass of sodium acetate, CH3COONa in the buffer.
pH = pKa + log [CH3COONa]/[CH3COOH]
=====> 4.9 = 4.75 + log [CH3COONa]/(0.1 M)
=====> 0.15 = log [CH3COONa]/[0.1 M)
=====> [CH3COONa]/(0.1 M) = antilog (0.15) = 1.4125
=====> [CH3COONa] = 1.4125*(0.1 M) = 0.14125 M
Since we have 100 mL buffer, hence, the mole(s) of CH3COONa present in the buffer is (100 mL)*(1 L/1000 mL)*(0.14125 M) = 0.014125 mole.
Molar mass of CH3COONa = (2*12.011 + 3*1.008 + 2*15.999 + 1*22.989) g/mol = 82.033 g/mol.
Mass of CH3COONa required = (0.014125 mole)*(82.033 g/mol) = 1.1587 g (ans).
ii) The buffer contains (100 mL)*(1 L/1000 mL)*(0.1 M) = 0.01 mole CH3COOH and 0.014125 mole CH3COONa.
NaOH reacts with CH3COOH in the buffer on 1:1 molar ratio.
Moles of CH3COOH in 40 mL buffer = moles of NaOH neutralized = (40 mL)*(1 L/1000 mL)*(0.1 M) = 0.004 mole.
Volume of 0.1 M NaOH neutralized by 40 mL buffer = (0.004 mole)/(0.1 M) = 0.04 L = (0.04 L)*(1000 mL/1 L) = 40 mL (ans).
iii) HCl reacts with CH3COONa in the buffer on a 1:1 molar ration.
Moles of CH3COONa in 40 mL of the buffer = moles of HCl = (40 mL)*(1 L/1000 mL)*(0.14125 M) = 0.00565 mole.
Volume of 0.1 M HCl neutralized = (0.00565 mole)/(0.1 M) = 0.0565 L = (0.0565 L)*(1000 mL/1 L) = 56.5 mL (ans).
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