Question

a 50.0 ml solution of ca(oh)₂ with an unknown concentration was titrated with 0.340 m hno₃. to reach the endpoint, a total of 28.4 ml of hno₃ was required. given that 0.00483 mol of ca(oh)₂ were present, what was the concentration of the initial ca(oh)₂ solution?

Explanation:

Step1: Recall molarity formula

Molarity ($M$) = $\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume in liters.

Step2: Convert volume to liters

The volume of $Ca(OH)_2$ solution $V = 50.0\ mL=50.0\times10^{- 3}\ L$.

Step3: Calculate molarity

We know $n = 0.00483\ mol$ and $V = 50.0\times10^{-3}\ L$. Then $M=\frac{0.00483\ mol}{50.0\times10^{-3}\ L}$.
$M = 0.0966\ M$

Answer:

$0.0966$