I'm completed and hopelessly stuck on this question and I don't know where to start the whole diluting and dissolving a mass in a volume part confuses me especially. Here's the q:
In an experiment to find the concentration of some dilute sulphuric acid, a student diluted it exactly 10 times to give a concentration suitable for titration. He made up some standard sodium hydroxide solution by dissolving 1.00g of it in 250cm^3 of solution. He then found that 25.0cm^3 of the sodium hydroxide solution required 23.5cm^3 of the diluted sulphuric acid for neutralization. Calculate the concentration of the original dilute sulphuric acid in moldm-3.
2NaOh + 2H2sO4 --> Na2SO4 + 2H2O
(Mr of Na = 23, and O= 16)
Titrations calculations help! Watch
- Thread Starter
- 01-01-2013 23:16
- TSR Support Team
- 01-01-2013 23:23
Right, firstly then calculate the number of moles of sodium hydroxide in the aliquot that was neutralized in the titration.
To do so, you calculate the amount of NaOH dissolved in the initial solution, then convert that to the amount in the solution taken e.g. if the moles in 250cm3 was 0.1 then the moles in 25cm3 would be 0.01.
Now, you've got the moles of NaOH that reacted.
So, now you calculate the number of moles of H2SO4 by their ratios in the balanced equation, to find the number of moles in 23.5cm3. Then, to find the moles in the initial volume, times by 10 since it was diluted ten times.
Now you've the number of moles in the initial volume, convert to concentration via the equation moles = concentration * volume.