3. If 46.53 mL of 1.500 M NaOH are required to titrate 25.00 mL of an unknown ac

Question

3. If 46.53 mL of 1.500 M NaOH are required to titrate 25.00 mL of an unknown acetic acid solution (vinegar). a.) how many moles of NaOH were required? Show work b.) How many moles are acetic acid are neutralized? c.) what is the molarity of the acetic acid? Show work. 4. The Ht ion concentrations in apricot, asparagus, and banana solutions are approximately 1.6 x 10 M, 2.5 x 10 9 M, and 2.5 x 10 M respectively. Determine the pH of apricot, asparagus, and banana solutions. 5. List the three foods of the previous question in order of decreasing acidity. 50

Explanation / Answer

3)
a) How many moles of NaOH were required?

Volume of NaOH solution V = 46.53 mL = 0.04653L

no. moles of NaOH = Molarity x volume in L
                   = 1.5 moles/L x 0.04653 L
                   = 0.0698 moles
                  
Therefore, no. moles required = 0.0698 moles

b) How many moles are acetic acid are neutralized?

CH3COOH + NaOH ---> CH3COONa + H2O

Since one mole of acetic acid reacts with one mole of NaOH; the no. moles of acetic acid = no. of moles of NaOH

No. moles of acetic acid neutralized = 0.0698 moles

c) What is the molarity of the acetic acid?

Volume of acetic acid V = 25 mL = 0.025L

Molarity of acetic acid = no. moles/ volume in L
                       = 0.0698 moles / 0.025 L
                       = 2.7918 M
                      
                       

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