Thymol blue in its acid range is not a suitable indicator for the titration of HCl by NaOH(aq) . Suppose that a student uses thymol blue by mistake in the titration of Figure 17-9 in the textbook and that the indicator end point is taken to be pH=2.0.
Approximately what percent of the HCl remains unneutralized at pH=2.0?
Well, if thymol blue is not a suitable indicator for the titration, then using it by mistake is like wearing a swim cap in the desert - it's just not gonna work out. But let's try to calculate this anyway.
Since thymol blue is in its acid range, it changes color from yellow to blue over the pH range of 1.2 to 2.8. So, if the indicator endpoint is taken to be pH=2.0, we can assume that at pH=2.0, the solution is still slightly acidic.
Now, we don't know the exact concentration of HCl or NaOH, so we'll have to make some general assumptions here. Let's say we start with 100 mL of HCl.
From the pH range of thymol blue, we can guess that around 99% of the HCl has been neutralized when the pH reaches 2.0. So, approximately 1% of the HCl remains unneutralized at pH=2.0.
Of course, this is just a rough estimate, and the actual percentage will depend on the specific concentrations and stoichiometry of the reaction. But remember, always choose the right indicator for titrations to avoid clown-like mistakes!
To determine the approximate percentage of HCl that remains unneutralized at pH=2.0 using thymol blue as an indicator, we need to consider the indicator's acidic range.
Thymol blue changes color within a pH range of 1.2 to 2.8. At pH values below 1.2, thymol blue appears yellow in color (indicating the presence of an acid), and above pH 2.8, it turns blue (indicating the presence of a base).
Since the student took the indicator's endpoint as pH=2.0, which is within the acidic range, it means that the solution still contains some HCl when the endpoint is reached.
Therefore, at pH=2.0, there will still be some unneutralized HCl present in the solution. The percentage of HCl remaining unneutralized can vary depending on the specific conditions of the titration. Unfortunately, without further information about the titration process, it is not possible to determine the exact percentage of HCl remaining unneutralized at pH=2.0.
It is important to note that using thymol blue, which has an acidic range that overlaps with the endpoint of the titration, is not suitable for accurately determining the percentage of HCl remaining unneutralized. A more appropriate indicator, such as phenolphthalein, should be used to indicate the complete neutralization of HCl by NaOH.
To determine the approximate percent of HCl that remains unneutralized at pH 2.0, we need to consider the pH range at which thymol blue changes color. Thymol blue is a pH indicator that transitions from yellow to blue within a pH range of 1.2 to 2.8.
However, it is important to note that the end point of an acid-base titration occurs when the moles of acid and base are stoichiometrically equivalent, which ideally should be at a pH of 7.0. In this case, the student mistakenly used thymol blue as the indicator and the pH 2.0 as the end point.
Since thymol blue changes color within a range of pH 1.2 to 2.8, we can approximate that at pH 2.0, the color change would be somewhere around the midpoint of the transition range. Therefore, the concentration of HCl at pH 2.0 is roughly halfway between the initial concentration of HCl and the concentration at the end point (pH 7.0).
To calculate the percent of HCl remaining unneutralized at pH 2.0, we need to compare the difference between the initial and final concentrations with the initial concentration:
Percent remaining unneutralized = [(Initial concentration - Final concentration) / Initial concentration] x 100
However, since we don't have specific concentration values, we cannot provide an exact percentage. We can only provide general information based on the principles.
It is worth noting that using thymol blue as an indicator for the titration of HCl by NaOH is not recommended because its pH transition range does not align well with the expected pH changes during the titration. A more suitable indicator, such as phenolphthalein or methyl orange, would be preferred for this titration.
I have no idea what concn is being titrated in Fig 17-9; however, if we consider it 100 mL of 0.1M HCl being titrated with 0.1M NaOH, we can ESTIMATE it this way.
You start with 0.1M and you want the concn of H^+ to be 0.01 (pH of 2) at the end; therefore, the concn must be reduced by about 0.09. We can make an approximate correction for the dilution by the titrate then as 0.09 M x (100/109 = about 0.0826M.
If we start with 0.1M we will still have 0.1M-0.0826M = 0.0174 M at pH = 2 and the percent is (0.0174/0.1)*100 = about 17%. That is an estimate only. The actual percent, for 0.1M, can be calculated as follows:
Assume we take 100 mL of 0.1M HCl (10 mmoles) and titrate with 0.1M NaOH. Let x = mL NaOH added.
([mmoles HCl - 0.1x)/(mL HCl + x)] = 0.01M
(10-0.1x) = (0.01)(100+x) and solve for x. If I didn't make a mistake x = 81.8 and we check to see if that is a good value.
10 mmoles HCl to start
-8.18 mmoles NaOH added
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1.82 mmoles HCl left untitrated.
(1.82 mmoles/10 initially)*100 = 18.2% not yet titrated so a significant error is made by using thymol blue. This percent will change as the concn changes; therefore, it your Figure 17-9 is not 0.1M, a different percent of HCl will remain untitrated at pH = 2.