48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 mL mol HNO3. Determine the molarity of the Ca(OH)2 solution.
To determine the molarity of the Ca(OH)2 solution, we can use the equation:
M1V1 = M2V2
Where:
- M1 is the molarity of the HNO3 solution,
- V1 is the volume of the HNO3 solution used during titration,
- M2 is the molarity of the Ca(OH)2 solution,
- and V2 is the volume of the Ca(OH)2 solution.
From the information given in the question:
- M1 = 0.385 mol/L
- V1 = 19.2 mL = 0.0192 L
- V2 = 48.0 mL = 0.0480 L
Plugging in these values into the equation, we have:
(0.385 mol/L)(0.0192 L) = M2(0.0480 L)
Simplifying the equation, we get:
0.007392 mol = 0.048 M2
Now, divide both sides of the equation by 0.048 to solve for M2:
M2 = 0.007392 mol / 0.048 L
Calculating this, we find:
M2 ≈ 0.154 mol/L
Therefore, the molarity of the Ca(OH)2 solution is approximately 0.154 mol/L.
Ca(OH)2 + 2HNO3 ==> Ca(NO3)2 + 2H2O
mols HNO3 = M x L = ?
Look at the coefficients. mols Ca(OH)2 = 1/2 mols HNO3.
Then M (Ca(OH)2 = mols/L = ?