A sample of solid monoprotic acid with molar mass equal to 169.7 g/mol was titrated with 0.1599 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.
Is the solution correct?:
moles NaOH = 0.1599 x 0.025=3.998x10^-3
moles H+ = moles NaOH (from equation) = 3.998x10^-3
mass H+ = moles x molar mass= 3.998x10^-3 x 169.7 = 0.6789g
I didn't check the math, but the procedure is correct.
The solution provided is correct. Here's how to calculate the mass of acid:
1. Calculate the moles of NaOH used: The concentration of NaOH solution is given as 0.1599 M, and the volume used is 25 mL (or 0.025 L). Multiply the concentration by the volume to obtain the moles of NaOH:
moles NaOH = concentration x volume
= 0.1599 M x 0.025 L
= 3.998 x 10^-3 moles
2. Since the acid and NaOH react in a 1:1 ratio, the number of moles of H+ ions produced is equal to the number of moles of NaOH used:
moles H+ = moles NaOH
= 3.998 x 10^-3 moles
3. Finally, calculate the mass of the acid by multiplying the moles of H+ ions by the molar mass of the acid. The molar mass of the acid is given as 169.7 g/mol:
mass H+ = moles H+ x molar mass
= 3.998 x 10^-3 moles x 169.7 g/mol
= 0.6789 g
Therefore, the mass of acid to be used is 0.6789 grams.