# Hardness of water using EDTA calculation issue

I have been asked to work out the hardness of water in German degrees and this is what I did:

I got a sample of water I didn’t measure how much I put into my volumetric flask perhaps it was $\pu{8.5ml}$ but I didn’t think I would need it, but I do use it which made me believe I was wrong.

The sample was made up to $\pu{100ml}$. $\pu{10ml}$ of this dilution was used as the analyte along with buffer and the indicator.

I Titrated $\pu{9.07ml}$ of EDTA with a concentration of $\pu{0.0185mol/dm^3}$

So the calculation I did was:

\begin{align} n(\ce{edta})&= (9.07\times0.0185)/1000= 0.000167795\\ n(\ce{Ca^2+}) &= n(\ce{edta})\end{align} So then I find the mass of Ca: $$m(\ce{Ca^2+}) = 40\times0.000167795= \pu{0.0067188g}$$

Now my issue:

So $\pu{0.0067188g}$ of Ca in the $\pu{10ml}$ therefore in the $\pu{100ml}$ there will be $10$ times that, as it was a dilution. So $\pu{0.067188g}$ which is $\pu{67.118mg}$.

So there is $\pu{67.118mg}$ of Ca in my $\pu{8.5ml}$ sample.
Now a $1^\circ\ \mathrm{dH}$ is $\pu{10mg}$ of Ca per $\pu{1000ml}$. So is there $\pu{7896.2mg}$ in $\pu{1000ml}$ of sample — therefore having $789.6^\circ\ \mathrm{dH}$ of hardness?