14

In acid-base titrations, synthetic indicators are exclusively used to find accurate end-point determinations because they always have a highly defined color change at certain pHs. For example, phenolphthalein ($\mathrm{p}K_\mathrm{a} = 9.7$ at $\pu{25 ^{\circ}C}$) is colorless in acidic solutions (precisely $0 \lt \mathrm{pH} \lt 8.2$), but it is pink in ...


8

The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$ that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}...


6

I'm not sure I follow your logic. For a monobasic acid S $(\ce{HA})$ dissociation degree $α$ is $$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$ where $c_\mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_\mathrm{b}$: $$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies ...


6

The main issue with natural indicators is their purity, long term chemical stability and shelf-life. Note that some natural indicators are pretty good for example the famous litmus paper is obtained from lichens. It is a mixture of several pigments like the red cabbage. Sometimes titrations are carried out at warm temperatures, do you think the pigments in ...


4

I'm not an expert on coordination chemistry, and in particularly, metal-indicator complexes in EDTA titrations for heavy metals. But, I assume, the addition of $\ce{[Cu(EDTA)]^2-}$ solution to your unknown $\ce{Pb}$ sample is to get a sharp end-point (example of Replacement or Substitution Titration). Since, the extra metal ion added ($\ce{Cu^2+}$) has ...


4

I have tried either inside or outside and they both worked fine. Since I deliver very small volumes at a time (20 μL), I validated the dispensed volume using a balance and dipping the tip inside the analyte was more precise. My system dispenses through a small micropipette tip (jerry-rigged).


4

In classical titrimetry you would avoid dipping the burette into the sample. However, for autotitrators, the electronic dispenser tip (electronic burette) should be dipped in the sample. This is to avoid any error due to the a drop clinging to the tip. In manual titration one would wash it. Most autotitrators are like this design including Karl Fischer ...


4

Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid. Unfortunately, this is wrong and not at all how stoichiometry works. Let's focus on what's important. In the nutshell, we are dealing with a typical neutralization reaction: $$\ce{H3O+ + OH- <=> 2 H2O}$$ and note that $\ce{BaCl2}$ as a ...


4

I believe have found a answer. It's valid mathematically, make sense physically but I don't know if chemically is true. I posted to community appreciation. There we go! The reactions ionization of weak acid: $$\ce{HA + H2O <=> H3O+ + A-}\qquad K_\ce{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}} \tag{1} \label{eq:KWeakAcid}$$ ionization of weak base: $$\...


4

Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point. The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount ...


4

You got the solubility part reversed. The solubility of $\ce{AgCl}$ is lower than the solubility of $\ce{Ag2CrO4}:$ $$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}$$ $$s(\ce{Ag2CrO4}) = \sqrt[3]{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt[3]{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$ ...


3

First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are: $$\ce{H2SO4(aq) + NaOH(aq) -> HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$ $$\ce{H2SO4(aq) + 2NaOH(aq) -> SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$ Your saying that reaction \eqref{rxn:2} might ...


3

Phenolphthalein becomes colorless again (from pink) in strongly basic medium with time. It is a known phenomenon called the alkaline fading of phenolphthalein. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=fading+of+phenolphthalein+in+alkaline+solution&btnG= It is not a good choice to use phenolphthalein in the presence of sodium ...


3

Easy marks for phenolphthalein usability are: $99.9\%$ of acetic acid is titrated at about $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{1000} = 7.75$$ $100\%$ of acetic acid is titrated at $$\mathrm{pH} = 7 + \frac12 \cdot (\mathrm{p}K_\mathrm{a} + \log c) = 9.385 + \frac12 \log c$$ Phenolphthalein color transition is $8.2-10.0.$


3

For the derivation of the Hendersson-Hasselbalch equation, it is assumed that both hydronium and hydroxide ions are minor species, i.e. there concentration is low compared to that of the buffer species. In your exercise, the hydronium ion concentration (according to your calculation) is about 0.03 M or higher. With a initial concentration of $\ce{H2SO3}$ of ...


3

In this case, cerium(IV) sulfate is a strong oxidising agent, hence, using $\ce{HCl}$ would likely be oxidised to form $\ce{Cl}$ of higher oxidation state. Therefore, the titre would be vastly higher than expected. $\ce{HNO3},$ on the other hand, is an oxidising agent and would oxidise $\ce{Fe}$ to a higher oxidation state than expected (of $\ce{Fe^3+}$). ...


3

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off. If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving ...


3

This practice is done in analytical chemistry in order to minimize the relative weighing error on the balance. Preferring a larger formula weight for a primary standard has nothing to do with impurity levels. We have to start with the highest purity standard. For example, you wish to prepare a 0.010 M solution of oxalic acid dihydrate in 1 L flask. Its ...


2

With respect to the reaction rate, the rate-limiting step of a titration is typically the mixing of the bulk solution. This is easy to see if you have an indicator dye in the solution. When you add a drop of the acid solution, there will be an immediate local coloring indicative of low pH (also demonstrating, as others said, that the proton transfer is ...


2

I think you might have a different understanding of what constitutes a 'change in the observation/reading' from the textbook, resulting in your confusion. I am assuming that the textbook is only considering the titration graph per se, pH vs volume of acid added. You are referring to the time to pH equilibrium during each drop of acid added. For the ...


2

The best approach is to ask the teacher again after doing some "homework" (i.e. reading and searching). Extracting and then titrating oxalic acid/oxalate is not that easy from spinach. You will need a lot of spinach to extract a titratable amount. Oxalic acid can be extracted by boiling spinach in 0.1 M HCl. However, tons of other redox active substances ...


2

The solute will alter the density of the solvent, which is used for example to determine the concentration of sulfuric acids by a hydrometer and in particular by help of a saccharometer. Depending on your equipment you will need at least about 20 mL of analyte. A popular alternative among winemakers is to rely on the change of the refractive index of a ...


2

If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point, $$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$ So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ ...


2

At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$. If we neglect $\ce{H+}$ from water dissociation and consider $c_\ce{A-}\ll c_\ce{HA}$, then $c_\ce{A-} = c_\ce{H+}$ Therefore $$K_\...


2

N-factor Is the change of atom oxidation state multiplied by number of atoms changing this state. Therefore, n-factor of $\ce{I2}$ is $2$, n-factor of $\ce{Na2S2O3}$ is $1$. So, equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass, for molecular iodine it is half of its molar mass.


2

Visually, we can only estimate the $\mathrm{p}K_\mathrm{a}$ value to be near the middle of the indicator range, shifted to more intense colour. The $\mathrm{p}K_\mathrm{a}$ of an indicator could be determined more precisely by combining $\mathrm{pH}$ meter with photometry, where we would get the well known " round step function" of absorbance versus $\...


2

I am afraid, direct acid-base titration is not the right way to analyze aspirin exactly because of the you stated - hydrolysis. The rule number no of any titration is that there should be no side reaction and it should go to completion almost instantly. The direct titration of aspirin is problematic because hydrolyzes pretty fast to salicylic acid- an ...


2

How a titration curve is affected when a poorly soluble salt is formed? If by "I think that titration curve don't be affected" you mean that the graph of pH vs volume of strong base will not be affected by the precipitations, then you are right. This is because the species that form the precipitate are not part of any other equilibrium reaction. how to ...


2

It is an interesting question. Daniel Harris is revising his book with my former mentor. Hope he clarifies this section in the revised version. Your point number 1 is misleading. The reason is that before the titration, theoretically there is no Fe(III). So Nernst equation should not be used- electrode potential is infinite (log 0 is undefined). When you ...


2

You cannot do simple volumetric redox titration to determine iron content because the amount of iron is very very small in a single fruit (on the order of fraction of a milligram). Imagine what would be the buret reading? Classical methods are good for large concentrations >> 1% wt/wt With such small quantities, UV-Vis absorption spectroscopy or atomic ...


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