If 20.0mL of an unknown acid was titrated with 2.0 M KOH and 60.0 mL KOH was used what is the molarity of the unknown acid?
If you don't know if the unknown acid was monoprotic, diprotic, or triprotic, you cannot determine molarity of the acid.
Assuming monoprotic (as in acetic acid, hydrochloric, nitric, citric)
Macid*Vacid=Mbase*Vbase
Macid=2M*60ml/20ml=6M
Normally, you should solve these equations using Normality as the concentration, but I suspect your teacher has not exposed you to that yet.
http://environmentalchemistry.com/yogi/chemistry/MolarityMolalityNormality.html
To find the molarity of the unknown acid, we can use the equation for the reaction between the acid and the base. In this case, the acid reacts with KOH to form water and a salt. The balanced equation for this reaction is:
Acid + KOH -> Water + Salt
Given that 2.0 M KOH was used, we can determine the number of moles of KOH used in the titration. The volume of KOH used is 60.0 mL, or 0.060 L. Therefore, the number of moles of KOH used can be calculated as:
moles of KOH = concentration of KOH x volume of KOH
= 2.0 M x 0.060 L
= 0.12 moles
According to the balanced equation, the ratio between KOH and the acid is 1:1. Therefore, the number of moles of the unknown acid is also 0.12 moles.
To find the molarity of the unknown acid, we need to calculate the concentration (molarity) using the equation:
Molarity = moles of solute / volume of solution in liters
In this case, the volume of the solution is 20.0 mL, or 0.020 L. Plugging in the values, we have:
Molarity = 0.12 moles / 0.020 L
= 6.0 M
Therefore, the molarity of the unknown acid is 6.0 M.