A student required 26.42 mL of 0.1013 M NaOH solution to titrate 0.154 grams of
ID: 550455 • Letter: A
Question
A student required 26.42 mL of 0.1013 M NaOH solution to titrate 0.154 grams of a solid triprotic acid to the phenolphthalein end point. Assuming she did her calcilations correctly, what did she report as the molar mass of her acid
Assuming you have the answer to this^ , please answer this Q:
The student later realized that she had read her buret incorrectly. The volume of the NaOH solution was actually 23.58 mL that she used. What kind of error (positive or negative) did this cause in her reported molar mass? Please explain.
Explanation / Answer
1 mol triprotic acid = 3 mol naOH
no of mol of NaOH reacted = 26.42*0.1013/1000 = 0.00268 mol
no of mol of triprotic acid = 0.00268/3 = 0.000893 mol
molarmass of triprotic acid = w/n = 0.154/0.000893 = 172.45 g/mol
if The volume of the NaOH solution was actually 23.58 mL that she used.
no of mol of NaOH reacted = 23.58*0.1013/1000 = 0.00239 mol
no of mol of triprotic acid = 0.00239/3 = 0.000797 mol
molarmass of triprotic acid = w/n = 0.154/0.000797 = 193.224 g/mol
so that, correct value is higher than first calculated,that results -ve error.
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