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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: olivianoelle on August 02, 2019, 06:22:42 PM

Title: pH calculations for acid/base titration help
Post by: olivianoelle on August 02, 2019, 06:22:42 PM

Hi,

Im trying to calculate the pH of my solution once I add a certain amount of base to it.

My acid is: 25mL .10190 M HCl + 50mL H2O

My base is: .1751 M NaOh

I need to calculate the pH after I add 7 mL of NaOH to the HCl solution, after 13.4 mL of NaOH is added, after 14.4 mL NaOH is added, and after 21.4 mL NaOH is added.

I did this calculation for after 7 mL NaOH is added. https://imgur.com/a/436qdlR (https://imgur.com/a/436qdlR)
My result came close to the pH observed actually in lab.

When I tried the same process with 13.4 mL NaOH added, I got a -log value of .44.

I dont know what to do :( Any help is appreciated. Thank you.

Title: Re: pH calculations for acid/base titration help
Post by: Borek on August 02, 2019, 06:32:50 PM

Hard to tell where the error is not seeing the incorrect calculations.

Title: Re: pH calculations for acid/base titration help
Post by: olivianoelle on August 02, 2019, 06:45:05 PM

Quote from: Borek on August 02, 2019, 06:32:50 PM

Hard to tell where the error is not seeing the incorrect calculations.

here are my wrong calculations along with the questions from the lab itself: https://imgur.com/a/IKWqlba (https://imgur.com/a/IKWqlba)

also here is my data: https://imgur.com/a/w5mLfsb (https://imgur.com/a/w5mLfsb)

Title: Re: pH calculations for acid/base titration help
Post by: Borek on August 03, 2019, 03:21:22 AM

No idea what you did with NaOH.

Title: Re: pH calculations for acid/base titration help
Post by: sunkal on August 03, 2019, 10:11:56 AM

olivianoelle,

Notice that 0.03397 is the molarity of the diluted HCl solution. It is not the # moles of HCl in the original acid solution taken in the titration (namely 25 mL of 0.1019 M). You are subtracting the # mol NaOH added from this 0.03397 which is wrong. You need to calculate the # mol HCl in 25 mL of 0.1019 M and from this you need to subtract the # mol NaOH added. Hoping it works out for you.

Title: Re: pH calculations for acid/base titration help
Post by: sunkal on August 03, 2019, 10:20:41 AM

Quote from: sunkal on August 03, 2019, 10:11:56 AM

olivianoelle,

Notice that 0.03397 is the molarity of the diluted HCl solution. It is not the # moles of HCl in the original acid solution taken in the titration (namely 25 mL of 0.1019 M). You are subtracting the # mol NaOH added from this 0.03397 which is wrong. You need to calculate the # mol HCl in 25 mL of 0.1019 M and from this you need to subtract the # mol NaOH added. Hoping it works out for you.

You have to redo your calculations also for the addition of 7 mL NaOH. The pH you got there was not correct either.

Title: Re: pH calculations for acid/base titration help
Post by: olivianoelle on August 03, 2019, 06:28:15 PM

Quote from: sunkal on August 03, 2019, 10:20:41 AM

Quote from: sunkal on August 03, 2019, 10:11:56 AM

olivianoelle,

Notice that 0.03397 is the molarity of the diluted HCl solution. It is not the # moles of HCl in the original acid solution taken in the titration (namely 25 mL of 0.1019 M). You are subtracting the # mol NaOH added from this 0.03397 which is wrong. You need to calculate the # mol HCl in 25 mL of 0.1019 M and from this you need to subtract the # mol NaOH added. Hoping it works out for you.

You have to redo your calculations also for the addition of 7 mL NaOH. The pH you got there was not correct either.

Thank you SO much for your help :)

Title: Re: pH calculations for acid/base titration help
Post by: sunkal on August 03, 2019, 09:45:39 PM

You are welcome. With pleasure.