Concentration Calculation

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Matthew Chan 1B
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Joined: Sat Sep 07, 2019 12:16 am

Concentration Calculation

Postby Matthew Chan 1B » Tue Oct 08, 2019 12:42 am

Can someone help me understand how I should go about solving this problem? Thanks.

Hard water is water that has high mineral content (mainly Calcium and Magnesium). A
concentration above 5.30 x 10^-3 M is considered to be very hard water. Assuming that
there are no magnesium ions present, is a 0.400 L solution of 0.120 g CaCO3 and
0.155g CaSO4 very hard water?

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Joined: Sat Aug 24, 2019 12:15 am

Re: Concentration Calculation

Postby Mashkinadze_1D » Tue Oct 08, 2019 1:11 am

To solve this problem we would do as follows:
1.20g CaCO3 x 1 mol CaCO3 /100.086g = 0.001199 mole CaCO3 or 1.20 x 10^-3 moles of CaCO3

0.155g CaSO4 x 1 mol CaSO4 / 136.134 g = 0.001138 mole CaSO4 or 1.14 x 10^-3 moles of CaSO4

( (1.20 x 10^-3) + (1.124 X 10^-3) )/ (0.400) = 0.00585 M or 5.85 X 10^-3 moles per liter

Yes, this is above the given level that would define this as hard water and therefore we can classify this as hard water.

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Joined: Tue Oct 08, 2019 12:16 am

Re: Concentration Calculation

Postby EricZhao3G » Wed Oct 09, 2019 1:48 pm

First, you would add the convert the grams of the 2 solutes to moles using the molar masses. Then, add those 2 moles to find the total moles of solute. Finally, divide the moles of solute by the volume of the solution in order to find the molarity and compare that to 5.30 x 10^-3 M to see whether or not it is hard water.

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