I had the opportunity to teach chemistry in a private school, I was to be their biology teacher since I was a graduate of human anatomy.

But on resuming for this term, the chemistry teacher travelled to South Africa and the school needed to replace her. They called me from my biology Lab to help them since the other guy considered for the job was a biology major graduate and couldn't handle chemistry syllabus.

To cut long story short I have a challenge with preparing for titration. Getting the quantity of acid to be mixed to get 0.050 mol/dm³ of H2SO4 and the amount of NAOH to neutralize the acid.

I would appreciate your help chemistry teachers on this platform. I have tried Google and YouTube videos no luck yet. Thanks in advance dear colleagues

Encoreunltd:
I had the opportunity to teach chemistry in a private school, I was to be their biology teacher since I was a graduate of human anatomy.

But on resuming for this term, the chemistry teacher travelled to South Africa and the school needed to replace her. They called me from my biology Lab to help them since the other guy considered for the job was a biology major graduate and couldn't handle chemistry syllabus.

To cut long story short I have a challenge with preparing for titration. Getting the quantity of acid to be mixed to get 0.050 mol/dm³ of H2SO4 and the amount of NAOH to neutralize the acid.

I would appreciate your help chemistry teachers on this platform. I have tried Google and YouTube videos no luck yet. Thanks in advance dear colleagues

Molar conc. = Mass conc. / Molar mass
Molar conc. = 0.05mol/dm^3
Molar mass = 98g/mol
therefore mass conc. 0.05 * 98 = 4.9g/dm^3

This means you'll you'll dissolve 4.9g of H2SO4 in 1dm^3 of water

Send a to me message privately...I will help out

Also, I have a job for you since you majored in human anatomy.

emorse:

Molar conc. = Mass conc. / Molar mass
Molar conc. = 0.05mol/dm^3
Molar mass = 98g/mol
therefore mass conc. 0.05 * 98 = 4.9g/dm^3

This means you'll you'll dissolve 4.9g of H2SO4 in 1dm^3 of water

Thanks a ton @emorse for your help

I did all of the above with 4 250mL tap water filled measuring cylinder but couldn't get a flash point with the NAOH solution and phenolphthalein indicator used. I thought maybe because I used tap water instead of distilled water.

Another possibility is the quantity of NAOH solution used. Am not just sure what is the problem is.

Encoreunltd:


Thanks a ton @emorse for your help

I did all of the above with 4 250mL tap water filled measuring cylinder but couldn't get a flash point with the NAOH solution and phenolphthalein indicator used. I thought maybe because I used tap water instead of distilled water.

Another possibility is the quantity of NAOH solution used. Am not just sure what is the problem is.

Prepare a 0.1mol/dm^3 solution of NaOH by dissolving 4g of it in 1dm^3 of water.
Also, use methyl orange indicator. It's more suitable for a strong acid - strong base titration.

Where are you chatting from?

Rursh:
Where are you chatting from?

Ikorodu, Lagos

emorse:

Prepare a 0.1mol/dm^3 solution of NaOH by dissolving 4g of it in 1dm^3 of water.
Also, use methyl orange indicator. It's more suitable for a strong acid - strong base titration.

Thanks for your time. Will get back to you with my findings tomorrow after practicals

...

emorse:

Prepare a 0.1mol/dm^3 solution of NaOH by dissolving 4g of it in 1dm^3 of water.
Also, use methyl orange indicator. It's more suitable for a strong acid - strong base titration.

Nailed it with methyl orange. The instructions requested for phenolphthalein indicator

Encoreunltd:


Nailed it with methyl orange. The instructions requested for phenolphthalein indicator

Phenolphthalein works better in weak acid - strong base titrations. Congrats by the way. And thanks for the feedback.