Astudent added 32.60 ml of 0.03020 m naoh to a 20.0 ml sample of ginger ale before his sample turned pink. (a) how many moles of naoh did he add to his solution? (b) how many moles of citric acid are in his sample? (c) how many grams of citric acid does this correspond to? (d) how many milligrams? how many mg/ml?

Solution:

molarity=

(a) Molarity of NaOH, = 0.03020 M

Volume of NaOH added,= 32.60 mL= 0.0326 L

(1L= 1000mL)

Number of moles of NaOH added :

0.03020 M NaOH means 1L of solution contains 0.03020 moles 0.0326 L contains = = 0.00098 moles

(b) Equal number of moles of acid  will neutralize equal number of moles of base, thus we can apply molarity equation:

Molarity of citric acid, =?

volume of citric acid, = 20 mL= 0.02 L

using formula, ,

= 0.049 M

(c) Grams of citric acid in 20 ml

moles of citric acid = of solution in Liters = = 0.00098 moles

Mass of citric acid =

= 0.1882 g

(d) 20 mL solution of ginger ale contains of citric acid

= 188.2 mg (1g = 1000mg)

Since, 20 mL solution of ginger ale contains of citric acid 188.2 mg

so 1 mL of solution contains =

                                              = 9.41 mg/ mL

the answer is d. 74.02