6B.5

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Jessica Tejero 3L
Posts: 54
Joined: Wed Sep 25, 2019 12:16 am

6B.5

Postby Jessica Tejero 3L » Thu Dec 05, 2019 9:12 pm

Hi, could someone help me approach this?

6B.5 Calculate the pH and pOH of each of the following aqueous
solutions of a strong acid or base: (a) 0.0146 m HNO3(aq); (b) 0.11 m
HCl(aq); (c) 0.0092 m Ba(OH)2(aq); (d) 2.00 mL of 0.175 m
KOH(aq) after dilution to 0.500 L; (e) 13.6 mg of NaOH dissolved
in 0.350 L of solution; (f) 75.0 mL of 3.5 3 1024 m HBr(aq) after
dilution to 0.500 L.

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

Re: 6B.5

Postby Osvaldo SanchezF -1H » Thu Dec 05, 2019 9:28 pm

You can start to approach this by simply knowing that the addition of pOH and pH Is equal to 14 so (pOH+pH=14). So if you find the pH of the molecule using the molarity you found in each problem you can just subtract it to 14 and get pOH or vice versa.

Jesse Anderson-Ramirez 3I
Posts: 54
Joined: Thu Sep 26, 2019 12:18 am

Re: 6B.5

Postby Jesse Anderson-Ramirez 3I » Sat Dec 07, 2019 2:14 pm

Osvaldo Sanchez Fernandez -4F wrote:You can start to approach this by simply knowing that the addition of pOH and pH Is equal to 14 so (pOH+pH=14). So if you find the pH of the molecule using the molarity you found in each problem you can just subtract it to 14 and get pOH or vice versa.

How do you solve for (c)?


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