If 39.2 of 0.177 is required to neutralize completely 22.0 solution, what is the molarity of the acid?
Is that 39.2 mL of 0.177 M that neutralizes 22.0 mL OF A MONOPROTIC BASE? What's with the no units thing? If the units I inserted are correct AND if the acid/base pairs have the same number of active hydrogen ions, then
mL x M = mL x M
To find the molarity of the acid, we need to first understand the given information.
We are told that 39.2 mL of a certain acid, when combined with 0.177 M (molar) sodium hydroxide (NaOH) solution, is required to neutralize completely the 22.0 mL of the sodium hydroxide solution.
To calculate the molarity of the acid, we can use the formula:
Molarity (M) = moles of solute / volume of solvent (in liters)
Step 1: Convert the volume of the acid solution (39.2 mL) and sodium hydroxide solution (22.0 mL) to liters:
Volume (L) = volume (mL) / 1000
Volume of acid = 39.2 mL / 1000 = 0.0392 L
Volume of sodium hydroxide = 22.0 mL / 1000 = 0.0220 L
Step 2: Calculate the number of moles of sodium hydroxide (NaOH) using its molarity and volume:
Moles = Molarity x volume (in liters)
Moles of NaOH = 0.177 M x 0.0220 L = 0.003894 moles
Step 3: Since the acid and sodium hydroxide react in a 1:1 ratio (acid to base), the moles of the acid would also be 0.003894 moles.
Step 4: Now, we can calculate the molarity of the acid by dividing the moles of acid by its volume:
Molarity of acid = moles of acid / volume of acid (in liters)
Molarity of acid = 0.003894 moles / 0.0392 L = 0.0993 M (rounded to four significant figures)
Therefore, the molarity of the acid is approximately 0.0993 M.