12.29d

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Jessica Lutz 2E
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

12.29d

Postby Jessica Lutz 2E » Mon Dec 04, 2017 11:08 pm

Calculate the pH and pOH of each of the following aqueous solutions of strong acid or base:
(d) 2.00 mL of 0.175 m KOH(aq) after dilution to 0.500 L
I understand the concept of finding pH and pOH but I am stuck on this question because I do understand how they got the molarity for the dilution. In the solutions manual they did (2.00ml/500ml)*.175mol/L to get the new molarity. I am confused about how this follows the rules of molarity.

Chem_Mod
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Re: 12.29d

Postby Chem_Mod » Tue Dec 05, 2017 12:55 am

This basically used M1V1 = M2V2. New concentration is M2 = M1V1/V2. V1/V2 represents your 2.00/500. M1 would be 0.175.

Yizhou Liu 3L
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Re: 12.29d

Postby Yizhou Liu 3L » Tue Dec 05, 2017 7:36 pm

You need to use the formula M1V1=M2V2 to get the new concentration of KOH, which also equals to the concentration of OH- because KOH is a strong base.


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