9

Much easier is to calculate the inverse function $[\ce{Na+}]=f([\ce{H+}],K_\mathrm{a1},K_\mathrm{a2},K_\mathrm{a3})$. Calculate fractions of respective phosphate forms as the function of $\ce{[H+]}$ and acidity constants. First, calculate the common denominator $CD$: $$a_1 = [\ce{H+}]^3$$ $$a_2 = [\ce{H+}]^2 \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{...


7

I am conducting a school-based research project into the antioxidative properties of green tea. Part of the procedure is to determine the concentration of antioxidant in green tea. This antioxidant is EGCG (a type of catechin). First of all, you cannot determine the concentration of just EGCG (epigallocatechin gallate) alone, because any extract of green ...


5

Titrations involving chelating of metal ions with ethylenediaminetetraacetic acid (EDTA) is an extremely useful analytical tool and a very well studied system in analytical chemistry. It can be bind to virtually any metal ion, evidently even to alkali metals. For instance, sodium ion complexes with EDTA in certain situations (Ref.1). The abstract of the ...


4

With a lot of assumptions, 0.5 M is a reasonable answer. Here are the holes you have to fill in. If it is a titration of NaOH and HCl, it is probably an acid/base titration where we stop the titration when the pH is neutral. Titrations are often performed to figure out the concentration of a solution. You would titrate a smaller sample of the solution to ...


4

Acid (Brønsted–Lowry Acid) is a substance that gives away its $\ce{H+}$, but the environment should be able to accept that $\ce{H+}$. After it accepted it - the acid (which is now its conjugate base) can accept it back. There are 2 extremes: The environment already has plenty of $\ce{H+}$ already. So even if acid gives its $\ce{H+}$ to the environment, it ...


4

The R code equivalent of Poutnik's answer, for refrence: # Sumamente facil Ka.1 <- 7.1 * 10^-3 Ka.2 <- 6.3 * 10^-8 Ka.3 <- 4.5 * 10^-13 Kw <- 10^-14 P_ca <- 1 pH.seq <- seq(from=0,to=14,length.out = 1000) a1 <- function(H) H^3 a2 <- function(H) H^2 * Ka.1 a3 <- function(H) H * Ka.1 * Ka.2 a4 <- function(H) Ka.1 * Ka.2 * Ka.3 ...


4

Let's say you are titrating a weak acid $\ce{H3A}$ with a strong base. $$\ce{H3A + OH- -> H2A- + H2O}$$ $$\ce{H2A- + OH- -> HA^{2-} + H2O}$$ $$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$ When all $\ce{H3A}$ reacts with $\ce{OH-}$ to form $\ce{H2A-}$, the first equivalence point is reached. Before the first equivalence point is reached, the $\ce{H3A}$ and $...


3

There are (at least) two reasons to consider: accuracy and precision. Accuracy as in preparing a stem solution of $\pu{100 mL}$ with a balance right to say $\pu{0.1 g}$ or a graduated cylinders right to the full $\pu{1 mL}$ is easier, than for $\pu{20 mL}$. Depending on the class, volumetric flasks may be better in terms of absolute / relative error than ...


3

Tips and hints would be greatly appreciated. My tips and hints for you (I apologize for if my tips and hints direct to me treating OP as a novice): Your dibasic acid is $\ce{HO2C-(CH2)n-CO2H}$. Thus, $\ce{A^2-}$ in the equation $(2)$ of Karsten Theis's answer is $\ce{^-OOC-(CH2)n-COO^-}$. Molar mass of this dibasic acid is: $2 \times (12+32+1) + 14n = 90 + ...


3

Let consider ratio of N/P. Let fertilizer A has this ratio N/P=a. Let fertilizer B has this ratio N/P=b. Let you want the ratio c, where a < c < b. Then you need to mix A : B in ratio $$\frac{ b - c}{c-a}$$ In the Central Europe with German scientific influence, it is called the "mixing cross rule". Note that you can chose just ratio of 2 ...


3

Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of $\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. ...


3

I will try to explain in the third case of disproportionation reaction first after that I believe you would be able to derive the first two cases easily. Let's start We will begin with the definition of n-factor. n-factor is the total number of mole of electron transferred for oxidation or reduction of 1 mole of molecule. Now let's take a general ...


2

You are missing an important point. Conductometric titration is not just plotting the conductance after adding the titrant. One cannot plot conductance as it is, because you are significantly diluting the solution. You have to apply a correction to take dilution into account. Once you apply a correction, the curve will remain flat because further addition ...


2

The reason is that strong acids have $\mathrm{p}K_\mathrm{a}$ values that are poorly known. These $\mathrm{p}K_\mathrm{a}$s cannot be determined directly, because the exact concentration of the ions is difficult to know with precision: the electrodes do not react with the concentration of $\ce{H^+}$ or $\ce{H3O^+}$ ions. They react with the activity of $\ce{...


2

The formula $$C_1 \times V_1 = C_2 \times V_2$$ should never ever be used for titrations calculations. It fails miserably in most of the cases (e.g. titration of sulfuric acid with NaOH). This formula only works when the concentrations are in normality, which is considered an obsolete unit in modern chemistry. The proper use of this relation is for ...


2

It is the molarity of total acetic acid/acetate content, so the 3rd option. It is quite general principle applied in the pH buffer context. You can have a wide rage of $\ce{pH}$ values with the particular buffer type, with the same total molarity of "active substance", like citrate, phosphate or mixed buffers, being adjusted by variable amount of ...


2

Real experiments and their outcomes are more interesting than expected so don't call it a trouble. First think of a couple of scenarios. a) If you just leave a pellets of solid NaOH in open atmosphere. It would absorb water and form a protective coating of insoluble sodium carbonate. If you analyze the solution, that would be still be very high in NaOH conc. ...


2

To use the formula that you indicate ($V_1 * C_1 = V_2 * C_2$), it is therefore necessary to know the volume of acetic acid $V_1$ or you only have the mass $m_1$. To switch from one to the other, you need to know / determine the density of the acetic acid used : you just need to take a volume $V$ of acetic acid and weigh it, the ratio $\frac{m}{V}$ ...


2

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$. I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of ...


2

A one loose fault of the question is not giving the nature of strong acid. Thus, it is safe to assume that $\ce{NaOH}$ is the strong base, and the overall chemical reaction can be written as: $$\ce{H2SO3 + 2 NaOH -> Na2SO3 + 2 H2O}$$ Thus, each $\pu{mol}$ of $\ce{H2SO3}$ need $\pu{2 mol}$ of $\ce{NaOH}$ to neutralize. You have correctly calculate $\pu{1....


2

Your second equation is an equilibrium. Both acetic acid and acetate ions are always present in a solution, although their proportion may vary. When $\ce{CH3COOH}$ is dissolved into water, a fraction of the molecules are transformed by reaction with water, producing some $\ce{H3O+}$ But if these ions are destroyed by adding $\ce{OH-}$ ions, more molecules $\...


1

The PH at equivalence point of titration for a strong acid and strong base should be 7 theoretically specking because, they produce a neutral salt on reaction. This means that the salt wont be hydrolysed or anything like that to raise the PH above 7. But during actual titration its important that we use an indicator to map the equivalence point. Generally ...


1

$\ce{CH2ClCOOH}$ is a weak acid; $\ce{NaOH}$ is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid? They react. $$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O} $$ A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base. $$N_1V_1=N_2V_2 $$ At this stage, an equivalence ...


1

If $\pu{1 l}$ of $\pu{1 M}$ $\ce{HCl}$ is gradually neutralized by adding $x\,\pu{mol}$ $\ce{NaOH}$ without change in volume, the $\mathrm{pH}$ of the obtained solution is given by $$\mathrm{pH} = -\log\left(\frac{1-x}{2} + \frac{1}{2}\sqrt{(1 - x)^2 + 4\times10^{-14}}\right).$$


1

I genuinely have no clue how to start a chemical equation with something like $\ce{HOOC(CH2)_nCOOH}$. Tips and hints would be greatly appreciated thanks. The only thing that an acid does in an acid/base reaction is to lose hydrogen ions. You can write the neutralization of a monoprotic acid with $\ce{NaOH}$ like this: $$\ce{HA(aq) + NaOH(aq) -> A-(aq) + ...


1

Since OP has productive conversation with Nicolas, I think the information extracted from chain of comments is good enough to convince OP has made fair enough effort to solve this homework. The following steps might be helpful as well: OP didn't indicate what $\ce{Zn/EDTA}$ molar ratio would this reaction take. As rule of thumb, it is generally $1:1$. Thus, $...


1

During titration of small amounts of acids, the molar amount of the indicator in 1-2 drops of $\pu{1 \%}$ indicator solution may not be negligible compared to the acid molar amount, affecting the result. So for that cases, $\pu{0.1 \%}$ solution is used, to be able to dose smaller indicator amounts. As the phenolphalein molar mass is about $M=\pu{318 g/mol}$,...


1

Yes, the reaction can further proceed to sulfur dioxide and sulfur trioxide, but only appears for certain, per my research, to occur in alkaline conditions. Here is a supporting opinion to this effect, which cites the corresponding reaction in alkaline conditions as: $\ce{3 S(2-) + 8 KMnO4 → 3 SO4(2-) + 8 MnO2 }$ that implies the further conversion of $\ce{...


1

For a weak acid $\ce{HA}$ with acid constant $K_\mathrm{a}$: $$\ce{HA + H2O <=> H3O+ + A-} \tag1$$ $$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} \ \Rightarrow \ [\ce{H3O+}] = K_\mathrm{a} \cdot \frac{[\ce{HA}]}{[\ce{A-}]} \tag2$$ Taking $-\log$ of both side of equation $(2)$: $$-\log [\ce{H3O+}] = -\log K_\mathrm{a} -\log \frac{[\ce{HA}]}{[\...


1

Let's calculate the $p$H by changing the volume of $\ce{HCl}$ added. I will use the unit millimole ($1 mmol = 0.001 mol$), which is more practical than the mole. In the very beginning the solution contains $50 mL~ \ce{NaOH}~ 0.1 M$, or $5 mmol$ $\ce{NaOH}$. Let's take, for example, the point when you have added $30 mL~ \ce{HCl} ~0.1 M$ (or $3.0 mmol$$ ˙\ce{...


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