A solution has a pOH of 8.20. What is the concentration of H3O+?

Chemistry

1 answer:

Allisa [31]3 years ago

3 0

The correct answer is B. $1.6\times 10^{-6}$

Explanation

The question provides us with information about the concentration of $[OH^{-} ]$ and we are supposed to use this information to find the $[H^{+} ]$ concentration. The pH of a solution gives us information about $[H^{+} ]$

If we the reaction is in equilibrium and with standard conditions of 25°C, then pH is given by the formula

$14 = pH + pOH\\\implies pH= 14 - pOH= 14-8.20= 5.8$ .

Once we find the pH we use the relationship $pH= -log[H^{+} ]$ , then the $[H^{+} ]$ ,is given by

$pH = -log[H^{+} ]\\\implies [H^{+} ]= 10^{-pH} = 10^{-5.8}= 1.58 \times 10^{-6}$

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If a molecule can hydrogen bond, does it guarantee that it will have a higher boiling point than a molecule that cannot? Explain

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Answer:

a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.

b): not necessarily due to London Dispersion Forces.

Explanation:

There are three major types of intermolecular interaction:

Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
Dipole-dipole interactions between all molecules.
London dispersion forces between all molecules.

The melting point of a substance is a result of all three forces, combined.

Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.

For example, $\rm H_2O$ (water) molecules are capable of hydrogen bonding. The melting point of $\rm H_2O$ at $\rm 1\; atm$ is around $0 \; ^{\circ}\rm C$ . That's considerably high when compared to other three-atom molecules.

In comparison, the higher alkane hexadecane ( $\rm C_{16}H_{34}$ , straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around $18\; ^{\circ}\rm C$ above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.

Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around $287\; \rm ^\circ C$ than that of HCl.