# The titration of 25 mL of a water sample required 15.75 mL of 0.0125 M EDTA. How do you calculate the hardness of water in the unit of ppm MgCO_3? (Assume the moles of EDTA are equal to the moles of MgCO3)

Dec 4, 2016

In $\text{parts per million}$ with respect to $M g C {O}_{3}$, the water is $17$ $\text{ppm}$.

We know that $\text{1 ppm}$ $\equiv$ $1 \cdot m g \cdot {L}^{-} 1$.

#### Explanation:

$\text{Moles of EDTA} = 15.75 \times {10}^{-} 3 \cdot L \times 0.0125 \cdot m o l \cdot {L}^{-} 1 = 1.97 \times {10}^{-} 5 m o l \cdot {L}^{-} 1$

And thus in a $1 \cdot L$ volume of this water, there are $1.97 \times {10}^{-} 5 m o l$ of $M g C {O}_{3}$, given the assumed 1:1 equivalence.

$\text{Mass of}$ $M g C {O}_{3}$ in a $1 \cdot L$ volume,

$=$ $1 \cdot \cancel{L} \times 1.97 \times {10}^{-} 5 \cdot \cancel{m o l \cdot {L}^{-} 1} \times 84.32 \cdot g \cdot \cancel{m o {l}^{-} 1} =$

$1.66 \times {10}^{-} 2 \cdot g$

With respect to $\text{magnesium carbonate}$, this is $17$ $\text{ppm}$, i.e. $17 \cdot m g$ of $M g C {O}_{3}$ $\text{per litre of solution}$.

Normally, we would use $\text{ppm}$ to express the concentration in terms of magnesium or carbonate ions separately in solution.

Note that when we deal with $\text{ppm}$ concentrations, typically we deal with trace quantities of ions in solution. And thus we don't really have to consider how the densities of these solutions vary. The densities are near as dammit to that of pure water.