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3 years ago

Why was einstein not happy with the probable positions of electrons in the orbitals?

1 answer:

6 0

Since electron orbitals are described as probability clouds, Einstein disagreement with the probable positions of electrons in the orbitals is that, It is not possible to know the orbit of an electron when the position is under probability.

According to Bohr's theory, it is difficult to locate electron or cannot be located in a definite region. Electron has to be found in an orbit and nowhere else. When the probability of finding an electron in a given spherical shell around the nucleus is plotted the distance of the electron from the nucleus for the hydrogen atom, the graph indicates that the probability of finding the electron increases as the distance between the electron and the nucleus decreases

Bohr claimed that electrons a entities had only probabilities if they weren't observed. While Einstein argued that they had independent reality.

But in wave mechanics Model, there is a slight chance of knowing the location of the electron.

Heisenberg uncertainty principle also claim the possibility of knowing the position of electron. Albert Einstein also claim that; to determine the position of an electron to an accurate extent, you would have to compromise your ability to know it's momentum. This inaccuracy will eventually affect the measurement of momentum which will be extremely uncertain.

Since electron orbitals are described as probability clouds, Einstein disagreement with the probable positions of electrons in the orbitals is that, It is not possible to know the orbit of an electron when the position is under probability.

Learn more here: brainly.com/question/2663406

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3 years ago

What is the pH of a mixture of equal volumes of 0.6 mM Lactic acid and 0.4 mM NaOH?

tino4ka555 [31]

Answer:

The pH of the mixture is approximately 4.2.

There's no equilibrium calculation involved if HCl is used instead of lactic acid. The pH of that solution will be approximately 3.7.

Explanation:

<h3>Lactic Acid</h3>

Lactic acid is a monoprotic acid. In other words, each lactic acid molecule can be ionized to release only one proton. Therefore, lactic acid will react with NaOH at a 1-to-1 ratio. The result will be one unit of the anion of lactate ion.

Lactic acid is in excess. All NaOH will be consumed. Before any dissociation takes place, the solution will contain 0.2 mM of lactic acid and 0.4 mM of lactate ions.

Lactic acid is a weak acid. The following equilibrium will take place in the solution:

$\rm C_3H_6O_3 \leftrightharpoons {C_3H_5O_3}^{-} + H^{+}$ .

Let the increase in the concentration of protons in the solution be $x\rm \;M$ . Construct a RICE table:

$\begin{array}{c|ccccc}\rm \textbf{R}&\rm C_3H_6O_3 & \leftrightharpoons &\rm {C_3H_5O_3}^{-}& +& \rm H^{+}\\\textbf{I} & 0.2\times 10^{-3} && 0.4\times 10^{-3} \\\textbf{C} & -x & & +x & & +x\\\textbf{E} & 0.2\times 10^{-3} -x & & 0.4\times 10^{-3} + x & & x\end{array}$ .

The pKa value of lactic acid is 3.86. Therefore, $K_{\rm a} = 10^{-3.86}$ . By the definition of the acid dissociation constant $K_{\rm a}$ ,

$\begin{aligned} K_{\rm a} &= \frac{\rm [H^{+}][{C_3H_5O_3}^{-}]}{[\rm C_3H_6O_3]}\\& = \frac{x(0.4\times 10^{-3} +x)}{(0.6 \times 10^{-3} - x)}\end{aligned}$ .

Given that $K_{\rm a} = 10^{-3.86}$ , solve for $x$ :

$\displaystyle \frac{x(0.4\times 10^{-3} - x)}{0.2 \times 10^{-3}-x} = 10^{-3.86}$ .

There might be more than one solutions. However, the correct solution shall ensure that all concentrations are greater than zero. In this case,

$x \approx 5.7\times 10^{-5}$ .

By the definition of pH,

$\rm pH = -\log_{10}{[H^{+}]} = -\log_{10}{5.7\times 10^{-5}} \approx 4.2$ .

The Henderson-Hasselbalch equation will yield a similar result.

Similar to lactic acid, HCl is also monoprotic. The solution will contain 0.2 mM of HCl and 0.4 mM of NaCl before any ionization takes place.

Unlike lactic acid, all that 0.2 mM of HCl will ionize to produce protons and chloride ions. The final proton concentration in the solution will be 0.2 mM. As a result,

$\rm pH = - \log_{10}{[H^{+}]} = -\log_{10}{0.2\times 10^{-3}} \approx 3.6$ .

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