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2 years ago

Which weak acid would be best to use when preparing a buffer solution with a ph of 9.20?

Chemistry

2 answers:

Murrr4er [49]2 years ago

8 0

Answer:

The acid with about $Ka=6.31x10^{-10}$ .

Explanation:

Hello,

By assuming that the dissociation of a weak acid is:

$H^+X^-H^++X^-$

From the following equation one computes pKa as long as the concentration of the conjugate base equals the concentration of the acid for this buffered solution:

$pH=pKa+Log(\frac{[Conj.Base]}{[Acid]})\\pKa=pH-Log(1)\\pKa=pH=9.2\\Ka=10^{-9.2}\\Ka=6.31x10^{-10}$

So the acid will be that with a $Ka=6.31x10^{-10}$ or close since the possible answers are not shown.

Best regards.

wariber [46]2 years ago

6 0

Answer: -

6.31 x 10⁻¹⁰

Explanation: -

The equation for buffer pH

pH = pKa + log $\frac{Conjugate Base}{Acid}$

If [Conjugate base] = [Acid], then preparing a buffer is best.

Then pH = pKa

pKa = - log Ka

Ka = $10^{-9.2}$

= 6.31 x 10⁻¹⁰

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Answer:

2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)

Explanation:

An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.

In this reaction, solid lithium reacts with liquid water to form soluble lithium hydroxide and gaseous hydrogen .

1. Molecular equation

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Lithium hydroxide is a soluble ionic compound, so we write it as hydrated ions.

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What are the limitations of litmus paper and Phenolphthalein indicators? Name to other indicators that can be used that do not h

jenyasd209 [6]

Answer:

Here's what I find.

Explanation:

An indicator is usually is a weak acid in which the acid and base forms have different colours. Most indicators change colour over a narrow pH range.

(a) Litmus

Litmus is red in acid (< pH 5) and blue in base (> pH 8).

This is a rather wide pH range, so litmus is not much good in titrations.

However, the range is which it changes colour includes pH 7 (neutral), so it is good for distinguishing between acids and bases.

(b) Phenolphthalein

Phenolphthalein is colourless in acid (< pH 8.3) and red in base (> pH 10).

This is a narrow pH range, so phenolphthalein is good for titrating acids with strong bases..

However, it can't distinguish between acids and weakly basic solutions.

It would be colourless in a strongly acid solution with pH =1 and in a basic solution with pH = 8.

(c) Other indicators

Other acid-base indicators have the general limitations as phenolphthalein. Most of them have a small pH range, so they are useful in acid-base titrations.

The only one that could serve as a general acid-base indicator is bromothymol blue, which has a pH range of 6.0 to 7.6.

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Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

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