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So, the type of indicator used for the experiment will overall determine the pH of the end point because some indicators turn a different
color in the presence of an acid and others in the presence of a base.
This is because when you are diluting a solution, you are affecting the volume of the solution, but not the number of moles present in the solution. So, in this experiment when 40cm3, 35 cm3 and 45 cm3 could have been used to dissolve the unknown acid and the number of moles would not be different.
180, the number of equivalents would have been two and not one, the number of moles of the unknown acid would have been 0.0090mol instead of 0.0045mol, and the molar mass of the acid would have been 220. Therefore, if the unknown acid had been diprotic everything would have been doubled.
In this experiment, an acid-base titration was used to determine the molarity of a NaOH solution, the number of moles of NaOH that reacted with a different unknown acid, and the molar mass of this unknown acid. This was done by making the concentrations of 0.10M HCl and NaOH equal to determine the molarity of NaOH which is 0.091M. We then found that 0.0045mol of NaOH reacted with a different unknown acid by using the molarity 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 use the same number of moles, 0.0045mol, for the acid to determine if molar mass. This was completed by using 0.0045mol and the mass in grams of the acid we used, which was 0.983g. By doing this we discovered that the molar mass of our acid was 220. By doing an acid-base titration
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