7th edition 6B number 3

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Megan_Ervin_1F
Posts: 78
Joined: Fri Sep 28, 2018 12:18 am

7th edition 6B number 3

Postby Megan_Ervin_1F » Fri Dec 07, 2018 7:56 pm

I don't understand why when this problem says 200ml of a .025M HCL solution you don't have to divide .025 by 200 to find the Molarity of H

Layal Suboh 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

Re: 7th edition 6B number 3

Postby Layal Suboh 1I » Fri Dec 07, 2018 10:04 pm

HCL is already in units of concentration (mol/L), so in order to find the number of moles, you would have to multiply by 200 mL (.2 L). In order to find the pH of the desired solution, you take the -log[.025 M]. In order to find the actual pH, you would need to divide the number of moles by .25 L and take the -log of that.


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