A 3.567 g sample of a monoprotic acid was dissolved in water. It took 15.24 mL of a 0.1578 M NaOH solution to?

Question

Not quite sure how to approach this equation at all.

Answer 1

to what? Are you looking for the molecular weight of the monoprotic acid? If so, remember M (molarity) is in moles/lter) and your colume of NaOH is in mL so you'ww want to convert that to liters.


.01524 liters x .1578 moles/liter = 0.002405 moles

so therefore you have 3.567 g / 0.002405 moles = 1483 gram/mole

use dimensional analysis and keep track pf the units to find the answer

Answer 2

The completed given information is that it took 15.24mL to neutralize the acid according to the equation HX + NaOH ===> NaX + H2O. The question is probably, "What is the molecular weight of the acid?"

The molecular weight has the units of gHX/moleHX. You have gHX in the given. The moles of NaOH can be turned into moleHX from the rest of the data.

Start with 3.567gHX/15.24mL NaOH. Multiply by 1000mL NaOH/0.1578mole NaOH. This comes from the molarity of the NaOH solution, 0.1578M. The mL NaOH cancel and gives you gHX/mole NaOH. Multiply that by 1mole NaOH/1moleHX. That comes from the balanced equation. The moleNaOH cancel and give you gHX/moleHX, which is the answer.