A student titrates 0.1719 g of an unknown monoprotic acid to the equivalence poi
ID: 535611 • Letter: A
Question
A student titrates 0.1719 g of an unknown monoprotic acid to the equivalence point with 21.85 mL of 0.1062 M NaOH. Which of the following is most likely to be the unknown acid? chlorous acid (MM 68.46 g/mol) propionic acid (MM 74.08 g/mol) nitrous acid (MM 47.01 g/mol) benzoic acid (MM 122.12 g/mol) lactic acid (MM: 90.08 g/mol) The molar solubility of calcium hydroxide (Ca(OH)_2) is 1.06 times 10^-2 M. What is K_sp for this compound? 8.81 times 10^-10 7.31 times 10^-7 1.38 times 10^-16 4.76 times 10^-6Explanation / Answer
13.
Assume that monoprotic acid is HA.
HA+ NaOH = NaA + H2O
Moles of NaOH = Molarity * volume in L
= 0.1062 * 0.02185
= 0.0023 moles
Moles of acid = 0.0023 moles
Because recation occurred in 1:1 ratio.
Number of moles = amount in g/ molar mass
Molar mass = amount in g / number of moles
= 0.1719 g/ 0.0023 mole
= 74.74 g/ mole
Thus the correct answer is propionic acid
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