Titration without the PH meter: Calculate the molarity of your unknown, assuming

Question

Titration without the PH meter: Calculate the molarity of your unknown, assuming it is monoprotic acid or monobasic base. Show all calculations for one titration with good precision including units and correct number of significant figures. Summarize only the results for the runs that you are using for the runs that you are using for your analyze concentration determination. Remember to account for any dilutions. Run 1 Run 2

Final burette reading (ml) 15.60 30.70

Initial (ml) 0.50 15.60

Volume added (ml) 15.10 15.10

Colour and intensity both light pink of end point

Explanation / Answer

Solution :-

CH3COOH + NaOH ---- > CH3COONa + H2O

Lets first calculate the moles of the NaOH used for the titration

Moles = molarity * volume in liter

            = 0.113 mol per L * 0.0151 L

           = 0.0017063 mol NaOH

Since the mole ratio of the NaOH and acetic acid is 1 : 1

So the moles of acetic acid reacted are same as moles of NaOH

So moles of acetic acid = 0.0017063 mol

Volume of vinegar used = 25.0 ml = 0.0250 L

So molarity of the acetic acid = 0.0017063 mol / 0.025 L

                                                        = 0.06825 M

concnetration is same for both tirals since same volume of NaOH used in both trials.

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