# Tag Info

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The thing with universal indicators is that you have 6 different colors or so differentiating certain pH-ranges. Now if you want to titrate to a certain pH-Value, it is easier to have an indicator like phenolphtalein that changes from colorless to a color at a certain point (phenolphtalein being colorless from 0 to 8 and red for $< 0$). So you can ...

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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 ...

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The used analytical reaction of purple permanganate to colourless $\ce{Mn^2+}$ occurs under acidic conditions: $$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$ Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide: $$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$ Therefore, sulfuric acid is added to make the solution ...

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Yes borax is a hydrate, but the reason it is favored for titration is that it gives consistent results, which to most scientists, is far more important than accuracy. The $\ce{ .\!10H2O}$ hydration number will remain constant and will not fluctuate to $\mathrm{11}$ or $\mathrm{9}$ within normal temperature and humidity ranges. This means that per mass the ...

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Without a proper closed container, sodium hydroxide readily attracts water and become a viscous solution (concentrated sodium hydroxide). This solution readily attracts carbon dioxide in air and reacts with it. $$\ce{CO2(g) + H2O(l)<=>H2CO3(aq)}$$ $$\ce{2OH^{-}(aq) + H2CO3(aq) -> 2H2O(l) + CO3^{2-}(aq)}$$ (The $\ce{OH^{-}}$ in the second equation ...

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When phosphoric acid is added to an aqueous solution of iron (III), the ligands in the yellow complexes $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$ ( $\ce{[Fe(H2O)6]^3+}$ are weakly colored but fairly acidic), are successively substituted by $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. This yields colorless hydrogenphosphatoiron(III) complexes, like $\ce{[Fe(... 10 If you look at the titration curve, which plots the volume of base added vs pH (source): you can see that the equivalence point occurs at pH = 7. Phenolphthalein is fuchsia in pH's roughly between 8.2 and 12, and is colorless below pH 8.2. When the number of moles of added base is equal to the number of moles of added acid (or vice versa; example valid for ... 10 There are some different applications. For one, it tells you how many times an acid can lose a proton. For instance you can have a diprotic ("with 2 H+") molecule, with n=2. For instance, H2SO4, deprotonates to HSO4-, which can then go to SO4(-2). Why is that important? Take an acidic reaction with H2SO4. If the extent of reaction is strongly forward, you ... 10 One thing that that has already been mentioned in the comments is the impact of$\ce{CO2}$on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid. The equilibrium between$\ce{CO2}$and$\ce{H2CO3}$$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on the pH of your aqeous system. As long as it is sufficiently ... 10 [...] the pipette tip should be well below the surface of the solution [...] Usually, one would indeed keep the pipette above the solution to avoid that a (minor) part of the solution forms a drop (or film) on the tip of the pipette and is carried away. Here, when the amount of oxygen dissilved in the solution is to be determined, another effect prevails. ... 9 Two kinds of reactions have to be considered: Neutralisation of the weak acid with the strong base Secondary reactions from the products The neutralisation is fairly straightforward: $$\ce{HA + NaOH -> NaA + H2O}$$ However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots ... 9 Prologue The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the ... 8 Using a little extra phenolphthalein won't matter. You just need enough so that the color is visible once the pH becomes basic enough. If you were titrating a large volume, you might need more than a couple of drops, but 2 drops is enough to be sure that the color will be visible in a typical lab-scale titration vessel. One drop would probably even work. The ... 8 In addition to Martin's answer, there is at least an old recommendation by E. B. Sandell and T. S. West in Pure Appl. Chem., 1969, 18, 427-436 (DOI), which states: Buffer capacity or buffer index. The capacity of a solution to resist changes in pH on addition of acid or base, which may be expressed numerically as the number of moles of strong acid or ... 8 For a much shorter and clearer explanation of this problem, look here. The Henderson-Hasselbalch (HH) equation is not to blame here. It is an approximate equation, with a certain region of validity. By its nature, it does not take into account the self-dissociation of water, which becomes increasingly important in dilute solutions. When concentrations reach ... 8 The bottom of the meniscus (in the middle) is your measurement point and should lie on the line that you are trying to read. This is completely dependent on the type of pipette that you are using. Some pipettes are designed so that you just leave the last drop and some you are supposed to just touch the tip of the pipette to the surface of what you've ... 8 I would be more worried about carbon dioxide (rather than water) being absorbed into the NaOH solution. A 1 molar NaOH solution is not going to be appreciably hygroscopic. But it will absorb CO2: two moles of hydroxide yield one mole of carbonate ion, which, depending on what you are titrating, may act as only deprotonate one mole of acid per mole base - ... 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:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}... 7 Outline There are three main points of interest in the titration graphic, and two you have found yourself. The three points are The start at V_\text{titr}=0 Halfway to the equivalence point at V_\text{titr}=10~\mathrm{mL} The equivalence point at V_\text{titr} = 20~\mathrm{mL} Why is number 2 important? Because that is where we can read out the \... 7 Molarity of diluted \ce {H2SO4} (solution 2): \pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M} (I may be using the wrong volume, is it possible that I have to add the \pu{25 ml} to the \pu{23.81 ml} and divide by \pu{0.04881 L}?) No, you used the correct volume since you want to know the concentration of the \pu{25 mL} that you added to the ... 7 Basically the law of equivalence wants you to balance the equivalents that are involved in your reaction. I'd like you to view you reactions in two parts, the oxidation and reduction halves. I see:$$\begin{align} \ce{I- -> I2} \tag{oxidation} \\ \ce{IO3- -> I2}\tag{reduction} \end{align}$$In other words, we've just discovered a (reverse ... 6 When using \ce{MnO4-} as an oxidant, pH is crucial. In acidic medium: \ce{Mn(VII) -> Mn(II)}$$\ce{8 H3O+ + MnO4^{-} + 5 e- -> Mn^{2+} + 12 H2O}$$Under neutral conditions: \ce{Mn(VII) -> Mn(IV)}$$\ce{2 H2O + MnO4^{−} + 3 e− -> MnO2 + 4 OH^{−}}$$In strongly alkaline medium: \ce{Mn(VII) -> Mn(VI)}$$\ce{MnO4^{−} + e− -> MnO4^{... 6 1. It all depends on the\text{p}K_{\text{in}}$of the indicator. If we look at the Henderson-Hasselbalch equation: $$\text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\ce{A-}]}{[\ce{HA}]} \right)$$ You can see that when$\text{pH}=\text{p}K_{\text{a}}$, then logarithm part must equal 0. This occurs when we have equal parts of the protonated and ... 6 The answer is A. Let's look at a weak acid: acetic acid. It has a$K_a$of$1.8\cdot 10^{-5}$. From this you can calculate how much it will dissociate in water. Let's say we have a concentration of$1~\mathrm{M}: \begin{align} K_a &= \frac{[\ce{CH3COO^{-}}][\ce{H+}]}{[\ce{CH3COOH}]}\\ 1.8 \cdot 10^{-5} &= \frac{x \cdot x}{1~\mathrm{M} - x}\\ x &... 6 An indicator works when the solution's pH is in its range. It's not necessarily the equivalence point. The point is called the end point where it actually changes color. If the difference in volume is small between the equivalence and end points, then the indicator can be used. Since a gradual change implies more volume must be added to obtain the same ... 6 0.12N is indeed 0.12 normal. In short, there are 0.12 moles of hydrogen ions per liter. Since this is sulfuric acid, it is 0.06 moles of sulfuric acid per liter. 0.12 moles of hydrogen ions per liter would give a pH just less than 1. To obtain a pH near 3, you would need to dilute 1 mL of this solution to 100 mL. For a pH of 4, dilute 1 mL to 1000 mL. 6 I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation): Let's consider the case of the titration of strong acid (volumeV_a$and concentration$C_a$: unknown) with a strong base (volume$V_b$and concentration$C_b$), and we'll denote$V_b(eq)$the volume of base at the equivalence: The pH increases slowly ... 6 To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand. Since we are working with dissolved$\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation: $$\ce{B + H2O<=>BH +OH-}$$ So the equilibrium ... 6 Upon researching this, I found that this a measurement of charge: $$\mathrm{cmol=10~\mathrm{meq}}$$ and $$1~\mathrm{eq}= \mathrm{F}$$ where$\mathrm{F}\$ is Faraday's constant and is: $$\mathrm{F}=96 485.3365~\mathrm{C}$$ So: $$\mathrm{cmmol}=\frac{\mathrm{cmol}}{1000}=\frac{\mathrm{eq}}{100000}=.96 4853365~\mathrm{C}$$ Sources: http://en.wikipedia....

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