Chemistry Forum

Find the percent acid (eq wt 173.80) if 20.07 mL of 0.1100 N base is required to neutralize 0.721 g of a sample.

What is the answer to this question?

The titration gives 20,07 ml * 0,11 mmol/ ml = 2,2077 mmol
Multiplied with the molar mass 173,8 mg/ mmol gives 383,6 mg = 0,3836 g
Divided to weight of sample gives 0,3836 g/0,721 g = 0,532 = 53,2%

Though, your answer looks to me correct, my doubt is if 173.80 g is equivalent weight, how can we take it as a "Molar mass"?

I think in this problem, we should assume (1equivalent acid/1 equivalent base)
20.07 mL x (1 L/1000 mL) x (0.1100 eq base/1 L) x (1 eq acid/1 eq base) x (173.8 g/1 eq) = 0.3837 g acid.

0.3837 g/ 0.721 g = 53.2%

It is the same result, what you get.

Check Definition of eq wt. Here as example for NaOH.

https://www.toppr.com/ask/question/the- ... f-naoh-is/