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3 years ago

Sketch the titration curve for the titration of a generic weak base B with a strong acid. The titration reaction is

Chemistry

1 answer:

Klio2033 [76]3 years ago

7 0

Answer:

Kindly check the attached picture.

Explanation:

Titration is one of the useful concept or practical in chemistry which is used in the determination of the concentration of a particular chemical specie. Without mincing words, let us dive straight into the solution to the question;

(a) Determining the equivalence point using the graph.

=> The equivalence point is the point at which there has been been "enough" addition of the acid to the base to give us the congugate acid. In the diagram showing the curve, it is the point labelled (a).

(b)The next thing is to determine the region with maximum buffering.

=> The position with the maximum buffering ranges from the point labelled (b) in the graph and it is the point when the acid and the base has been added. But the point with the maximum buffering is labeled as point (c).

(c). Determination of the point where pH = pKb.

=> The Point in which the pH = pKb is the point a

(d). The region where pH depends only on [B].

=> The starting point in the point in which the pH is dependent on the concentration of the base, [B].

(e). The region where pH depends only on [BH+].

=> This is the same point with point labelled as (a). Thus, point (a) and (e) are the same.

(f). The region where pH depends only on the amount of excess strong acid is the point labelled as (f) which is known as the post equivalence point.

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How much sulphur dioxide is produced on complete combustion of 1 kg of coal containing 6.23percent sulphur?

Tresset [83]

Answer:

Approximately $1.24 \times 10^{2}\; \rm g$ , assuming that all sulfur in that coal was converted to $\rm SO_2$ .

Explanation:

Look up the relative atomic mass of $\rm S$ and $\rm O$ on a modern periodic table:

$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

Convert the unit of the mass of coal to grams:

$\begin{aligned} & m(\text{coal})= 1\; \rm kg \times \frac{10^{3}\; \rm g}{1\; \rm kg} = 1000\; \rm g\end{aligned}$ .

Mass of sulfur in that much coal:

$m(\text{sulfur}) = 1000\; \rm g \times 6.23\% = 62.3\; \rm g$ .

The relative atomic mass of sulfur is $32.06$ . Therefore, the mass of each mole of sulfur atoms would be $32.06\; \rm g$ . Calculate the number of moles of atoms in that $62.3\; \rm g$ of sulfur:

$\begin{aligned}& n(\text{S})\\&= \frac{m(\mathrm{S})}{M(\mathrm{S})}\\ &= \frac{62.3\; \rm g}{32.06\; \rm g \cdot mol^{-1}} \approx 1.94323\; \rm mol\end{aligned}$ .

Each $\rm SO_2\!$ molecule contains one sulfur atom. Therefore, assuming that all those (approximately) $1.94323\; \rm mol\!$ of sulfur atoms were converted to $\rm SO_2$ molecules through the reaction with $\rm O_2$ , (approximately) $1.94323\; \rm mol$ of $\!\rm SO_2$ molecules would be produced.

Calculate the mass of one mole of $\rm SO_2$ molecules:

$\begin{aligned}& M(\mathrm{SO_2})\\ &= 32.06 + 2\times 15.999 \\ &= 64.058\; \rm g \cdot mol^{-1}\end{aligned}$ .

The mass of that $1.94323\; \rm mol$ of $\rm SO_2$ molecules would be:

$\begin{aligned}& m(\mathrm{SO_2}) \\ &= n(\mathrm{SO_2}) \cdot M(\mathrm{SO_2}) \\ &\approx 19.4323\; \rm mol \\ &\quad \times 64.058\; \rm g \cdot mol^{-1} \\ &\approx 1.24\times 10^{2}\; \rm g\end{aligned}$ .

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3 years ago

Which one of the following sets of units is appropriate for a second-order rate constant?A. s-1B. mol L-1 s-1C. L mol-1 s-1D. mo

Korolek [52]

Answer:

Explanation:

A second order reaction is one in which the reaction rate is proportional to the concentrations of the 2 reactants. We only consider the reactants whose concentration are changing however.

For example, in the case:

A + B ----> C + D , provided that both A and B undergo concentration changes, the rate is proportional to the product of their concentrations.

Now, like we know, the unit of the reaction rate is per second I.e 1/s or simply s^-1 . Also it is important to note that the concentration of the reactants is exprested in units of mol/L. We can now mathematically calculate the unit of the rate constant.

Mathematically,

R = k × [A] [B]

Hence,

S^-1 = k × (mol/L)(mol/L)

s^-1 = k × (mol/L)^2

K = s^-1 ÷ (mol/L)^2

K = s^-1 × L^2 × mol^-2.

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3 years ago

Hydrocarbons contain only _____. carbon and hydrogen carbon, hydrogen, and oxygen carbon and nitrogen carbon and helium

Sati [7]

Hydrocarbons contain only Carbon and hydrogen.

Hope this helps!