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volumeacid*molarity acid=volumebase*molarity base
MolarityHCl= 45*.250M/12.3
MolarityHCl= 45*.250M/12.3
M1V1 = M2V2
where M1 and V1 are the molarity and volume of the potassium hydroxide (KOH) solution, and M2 and V2 are the molarity and volume of the hydrochloric acid (HCl) solution.
In this case, you have the following information:
- M1 (Molarity of KOH) = 0.250 mol/L
- V1 (Volume of KOH) = 45.0 mL
- V2 (Volume of HCl) = 12.3 mL
To find M2 (Molarity of HCl), you need to rearrange the formula as follows:
M2 = (M1V1) / V2
Now substitute the given values into the equation:
M2 = (0.250 mol/L * 45.0 mL) / 12.3 mL
Calculate the result:
M2 = 1.147 mol/L
Therefore, the molarity of the HCl solution is 1.147 mol/L.