A of vinegar to a phenolphthalein end point. volume of 32.45 mL of 0.122 M NaOH

Question

A of vinegar to a phenolphthalein end point. volume of 32.45 mL of 0.122 M NaOH was required to titrate 4.89 g a. Calculate the moles of NaOH added. b. Calculate the moles of acetic acid in the vinegar sample. Recall c. Calculate the grams of acetic acid in the vinegar sample. d. Calculate the percent by weight acetic acid in the vinegar. that NaOH and acetic acid react in a 1:1 mole ratio. Show mathematically that, for the special case nn2 n, the equation for computing the pooled variance (equation 4) reduces to the form shown in equation 6. That is, derive equation 6 fron equation 4.

Explanation / Answer

Ans 1

a) Molarity = no. of moles / volume of solution in L

n = M x V

n = 0.122 x 0.03245

n = 0.0039589 moles

So the number of moles of NaOH will be 0.0039589.

b) Since the reaction has equimolar concentrations of both acid and base , so the number of moles of acetic acid will also be the same , i.e. 0.0039589 moles

c) The molar mass of acetic acid is 60.05 g/mol

So the amount in grams = 60.05 x 0.0039589

= 0.2377 g

d) Perecnt by weight = (0.2377 / 4.89) x 100

= 4.86 %

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