1. When the terminal point is reached in an acid-base titration. the relationship between the concentrations of OH- and H3O+ are that they are equal. This doesn’t mean that the pH will be impersonal. but the concentrations of both will be the same in a titration.
2. The pH of the terminal point is determined by what sort of index is used to bespeak when the terminal point I reached. In this experiment. phenolphthalein was used. It shows a colour alteration in the presence of a base. which means that our solution had to be somewhat basic for it to turn a pinkish-purple colour. So. the type of index used for the experiment will overall find the pH of the terminal point because some indexs turn a different
colour in the presence of an acid and others in the presence of a base.
3. Regardless of the sum of H2O that is used to fade out the unknown acid. the sum of moles would non alter. This is because when you are thining a solution. you are impacting the volume of the solution. but non the figure of moles present in the solution. So. in this experiment when 40cm3. 35 cm3 and 45 cm3 could hold been used to fade out the unknown acid and the figure of moles would non be different.
4. If the unknown acid ad been diprotic. so the mole-to-mole ratio between the acid and NaOH would hold been 2:1. the molar concentration and normalcy would hold been 0. 180. the figure of equivalents would hold been two and non one. the figure of moles of the unknown acid would hold been 0. 0090mol alternatively of 0. 0045mol. and the molar mass of the acid would hold been 220. Therefore. if the unknown acid had been diprotic everything would hold been doubled.
In this experiment. an acid-base titration was used to find the molar concentration of a NaOH solution. the figure of moles of NaOH that reacted with a different unknown acid. and the molar mass of this unknown acid. This was done by doing the concentrations of 0. 10M HCl and NaOH equal to find the molar concentration of NaOH which is 0. 091M. We so found that 0. 0045mol of NaOH reacted with a different unknown acid by utilizing the molar concentration of NaOH and the volume of NaOH that we used to titrate with the unknown acid.
Since the mole-to-mole ratio of NaOH and the unknown acid were 1:1. we could utilize the same figure of moles. 0. 0045mol. for the acid to find if molar mass. This was completed by utilizing 0. 0045mol and the mass in gms of the acid we used. which was 0. 983g. By making this we discovered that the molar mass of our acid was 220. By making an acid-base titration