Titration: Calculate Acid Mass for NaOH 25 mL

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SUMMARY

The discussion focuses on calculating the mass of a monoprotic acid required for titration with a 0.1608 M sodium hydroxide (NaOH) solution, specifically for a volume of 25 mL. Given the molar mass of the acid at 179.4 g/mol and a 1:1 mole ratio, the calculation shows that 25 mL of NaOH contains 4.020 x 10^-3 moles, leading to a required mass of 0.7212 grams of acid. The approach taken is confirmed as correct, with minor adjustments noted in the mole calculation.

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Homework Statement



A sample of solid monoprotic acid with a molar mass equal to 179.4 g/mol was titrated with 0.1608 M sodium hydroxide solution. Calculate the mass in grams of acid to be used if the volume of NaOH to be used is 25 mL.

The Attempt at a Solution



Since the acid is monoprotic the ratio is 1:1, thus the number of moles in 25 mL of NaOH is the same number of moles of acid needed. The mass in grams can then be calculated using the molar mass given.

From this aproach:
25 mL NaOH contains 4.020 *10^-3 mol NaOH
4.020 *10^-3 mol acid = 4.020 *10^-3 mol * 179.4 g/mol = 7.212 *10^-1

Is this the right approach or am I going in the wrong direction?
 
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4.20 not 4.02, but you are on the right track.

Does the question explicitley state "volume to be used is 25 mL" or does it rather say something about burette being 25 mL?
 

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