If 31.00 mL of a dilute solution of lime water (calcium hydroxide) required 10.10 mL of 0.246 M hydrochloric acid solution for neutralization to a methyl red end point, calculate the molarity of the lime water.
Can you please explain how to do it?
Do I use M1V1=M2V2?
M1V1 = M2V2 is a shortcut to these problems when they are 1:1 mole ratio but ONLY then. This one is not so you must go the long way.
Write the equation.
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
moles HCl = M x L = ??
Use the coefficients in the balanced equation to convert moles HCl to moles Ca(OH)2.
moles HCl x (1 mole Ca(OH)2/2 moles HCl) = ??moles HCl x (1/2) = xx moles Ca(OH)2.
M Ca(OH)2 = xxmoles Ca(OH)2/L Ca(OH)2.
Thanks soooo much!!!
Yes, to solve this problem, you can use the equation M1V1 = M2V2, also known as the dilution formula.
First, let's identify the relevant information given in the question:
V1 = volume of lime water solution = 31.00 mL
M2 = molarity of hydrochloric acid solution = 0.246 M
V2 = volume of hydrochloric acid solution = 10.10 mL
Now, we need to find M1, which is the molarity of the lime water solution.
To do this, we will rearrange the dilution formula to solve for M1:
M1 = (M2 * V2) / V1
Substituting the given values into the dilution formula, we have:
M1 = (0.246 M * 10.10 mL) / 31.00 mL
Now, let's calculate this:
M1 = (2.5066 mmol) / 31.00 mL
M1 = 0.0809 M
Therefore, the molarity of the lime water solution is 0.0809 M.