Analysis of Vinegar Preparation of a Standard Sodium Hydroxide Solution We begin

Question

Analysis of Vinegar Preparation of a Standard Sodium Hydroxide Solution We begin by diluting 6 M NaoH with water since diluting NaoH provides only an approximate concentration and it is necessary to know the concentration of NaoH precisely, we will prepare a standard solution by titration. First, we weigh crystals of potassium hydrogen phthalate, KHCsH40. (abbreviated KHP). After dissolving the KHP crystals in water, we will trate the acid solution with NaOH according to the following equation. KHP(aq) Naohl(aq) KNaP(aq) H2O(l) Example Exercise 1 Molar Concentration of Standard NaoH A0.515 g sample of KHP (204.23 g/mol) is dissolved in water and requires 12.75 mL of NaoH solution to reach a faint pink endpoint. Find the molarity of the NaoH solution. Solution ng to the equation for the reaction and applying the rules of Referri stoichiometry, we have I mel KHR. I mol NaOH 0.515 *HP moFRAF 0.00252 mol NaOH The molarity of the NaoH is found as follows: 0.00252 mol NaOH 000 mt. 0.1983 mol NaOH TC solution 0.198 M NaOH In this example, the concentration of the standard NaOH solution is 0.198 M.

Explanation / Answer

Given that the molarity of NaOH = 0.198 M

The reaction between CH3COOH + NaOH is as follows:

CH3COOH + NaOH = CH3COONa + H2O

For trail 1

Volume of Vinegar = 10 .0 ml

Volume of NaOH used = final volume – initial volume

= 24.50 ml-9.75 ml

= 14.75 ml

Now calculate the moles of NaOH:

Moalrity * volume in L

0.198 mole /L* 14.75/1000

= 2.9*10^-3 moles

This number of moles of acid is equal to the number of base.

Moles of CH3COOH = 2.9*10^-3 moles

Volume of CH3COOH = 10 .00 ml = 0.010 L

Molarity = number of moles / volume in L

= 2.9*10^-3 moles / 0.010 l

= 0.2905 M

= 0.3 M

Given that the density of vinegar = 1.01 g/ ml

Density = mass / volume

Mass = density * volume

= 1.01 g/ ml*10 ml

= 10.1 g

Mass of CH3COOH in vinegar = number of moles * molar mass

= 2.9*10^-3 moles *60.05 g/mol

= 0.174 g

Mass of vinegar = 10.1 g

% of CH3COOH =mass of acetic acid / mass of vinegar *100

= 0.174 g /10.1 g *100

= 1.72%

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