A student titrates a solution containing 0.254 g of a monoprotic acid in 35 mL o
ID: 487757 • Letter: A
Question
A student titrates a solution containing 0.254 g of a monoprotic acid in 35 mL of water. The student reaches the endpoint after adding 18.35 mL of 0.107 M NaOH. How many moles of the acid are present in the sample? What is molecular weight of the acid? Imagine that you are attempting to determine the concentration of an acid by titration with a solution of sodium hydroxide with a known concentration, using phenolphthalein as an indicator. For each of the following sources of error, indicate whether the calculated concentration of acid would be higher or lower than the true value. You added too much base and went beyond the end point. Answers without explanation will not be given any credit Circle the correction answer: The calculated concentration of acid would be than the true value. HIGHER LOWER Reason: _____ Some of the acid splashed on to the sides of the flask and was not titrated. Explain your answer. Answers without explanation will not be given any credit. Circle the correction answer: The calculated concentration of acid would be than the true value. HIGHER LOWER Reason _____Explanation / Answer
Base added = 18.35 ml of 0.107 M NaOH
Therefore, from the formula
M1V1 = M2V2
M1 = x , V1 = 35ml
M2 = 0.107 M , V2 = 18.35 ml
(x)(35) = (0.107)(18.35)
x = (0.107 * 18.35) /35 = 0.056 M
Molarity = moles/ volume(litres)
Moles = 0.056 * (35/1000) = 1.96 * 10^-3 mol ---à ans (a)
Moles = given weight/ molecular weight
Molecular weight = 0.254/(1.96 *10^-3) = 129.59 g. ---à ans (b)
a) too much base if added, Vb >Va
(Reason)
MaVa = MbVb
Ma = (MbVb)/ Va
(Ma/ Mb) = (Vb/Va) > 1
Hence the calculated concentration of acid will be higher than the true value.
b) If acid is splashed then the acid will be wasted (not titrated), Va < Vb
(Reason)
If acid is wasted according the base will also be adjusted
(Ma/Mb) = (Vb/Va)
Ma = (MbVb)/Va
Hence the calculated concentration of acid will be higher than the true value.
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