French Toast Number 27 Part B

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KateCaldwell 1A
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Joined: Fri Apr 06, 2018 11:02 am

French Toast Number 27 Part B

Postby KateCaldwell 1A » Tue Jun 12, 2018 9:03 pm

Part a has us solve for the volume, which I got correctly, but I keep getting the wrong pH value for part b. Can someone explain step by step how to do this? I did moles/volume the -log that value.

Chem_Mod
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Re: French Toast Number 27 Part B

Postby Chem_Mod » Tue Jun 12, 2018 9:18 pm

Metal oxides, like Na2O, form O2- in water. O2- reacts with water to form 2 moles of OH-. Therefore, you must make sure you double the molarity since each mole of Na2O yield 2 moles of OH-. Remember that you need hydronium and/or hydroxide ion concentrations to determine pH, not the concentration of the actual base or acid.

Gisselle Sainz 2F
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Re: French Toast Number 27 Part B

Postby Gisselle Sainz 2F » Wed Jun 13, 2018 12:52 pm

After doubling the molarity, I still didn't get the right answer.

Molarity= (.550/.21)= 2.62 x 2 = 5.24

and 14 minus the negative log of that doesn't result in the correct answer. I'm not sure what's wrong.

Lily Emerson 1I
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Re: French Toast Number 27 Part B

Postby Lily Emerson 1I » Wed Jun 13, 2018 11:31 pm

you just need to do 14-(-log(1.1)) since you already know that the concentration of the new mixture is .55 - just multiply that by 2 moles of HO- to get the concentration

Cindy Nguyen 1L
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Re: French Toast Number 27 Part B

Postby Cindy Nguyen 1L » Thu Jun 14, 2018 3:07 am

Lily Emerson 1I wrote:you just need to do 14-(-log(1.1)) since you already know that the concentration of the new mixture is .55 - just multiply that by 2 moles of HO- to get the concentration


Where did you get 1.1 from?


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