Titration: Calculate Acid Mass for NaOH 25 mL

AI Thread Summary
To calculate the mass of a monoprotic acid needed for titration with 25 mL of 0.1608 M NaOH, the molar ratio is 1:1. The moles of NaOH in 25 mL equate to approximately 0.004020 moles, which is also the amount of acid required. Using the provided molar mass of the acid (179.4 g/mol), the mass of acid needed is calculated to be about 0.7212 grams. There is a clarification on whether the volume specified is for the NaOH or related to a burette measurement. The approach taken is generally correct, with a minor correction 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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