Would the concentration of hydroxide ion in a 0.30 m solution of potassium hydroxide (koh) be greater than, less than, or equal to 0.30 m?
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Potassium hydroxide KOH is a strong base which can completely ionize to form potassium and hydroxide ions.
The equation representing ionization of KOH,
The mole ratio of hydroxide to KOH from the above equation is
The given concentration of KOH solution=0.30M
Calculating the concentration of hydroxide ions present in 0.30M KOH solution:
Therefore the concentration of hydroxide ion will be equal to 0.30 M in a 0.30 M KOH solution.
Final answer:
The concentration of hydroxide ion in a 0.30 m solution of potassium hydroxide would be equal to 0.30M. Potassium hydroxide is a strong base, it completely dissociates in water producing an equal concentration of hydroxide ions. In a basic solution where the concentration of Hydroxide ions is high, the concentration of hydronium ions will be low.
Explanation:
The concentration of hydroxide ion in a 0.30 m solution of potassium hydroxide (KOH) would be equal to 0.30 M. This is based on the fact that potassium hydroxide is a strong base and it completely dissociates in water, which means one KOH molecule produces one hydroxide ion. Therefore, the concentration of the hydroxide ion [OH-] in the final solution would be equal to the initial concentration of the potassium hydroxide.
This is different from the behavior of weak acids and bases which do not completely dissociate in water, leading to a lower concentration of hydronium (H3O+) or hydroxide (OH-) ions compared to the initial concentration of the acid or base.
However, it's important to note that in solutions with a high concentration of hydronium ions, the concentration of hydroxide ions will decrease according to Le Chatelier's principle, which describes how equilibrium adjusts when it's disturbed. Conversely, in a basic solution—like this 0.30 M KOH solution—where the concentration of hydroxide ions is high, the concentration of hydronium ions will be comparatively low.
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b. secondary succession has soil
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Hydrogen bonds between water molecules occur because of the polarity of the water molecule.
A hydrogen bond is a weak bond that forms between a hydrogen atom that is covalently bonded to a highly electronegative atom and another highly electronegative atom in the same or a different molecule.
A water molecule consists of two hydrogen atoms and one oxygen atom. The hydrogen atoms are bonded to the oxygen atom, but the electrons in the covalent bonds are not shared equally. This results in a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms.
When two water molecules are close together, the partial positive charge on one hydrogen atom is attracted to the partial negative charge on the oxygen atom of the other molecule. This attraction is called a hydrogen bond.
Therefore, due to polarity of the water molecule, it results into hydrogen bonds between water molecules.
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You would need to mix 0.100 liters (or 100 mL) of the 15M NaNO3 solution with enough solvent (e.g., water) to bring the total volume to 250 mL in order to obtain a 6.0M NaNO3 solution.
To prepare 250 mL of a 6.0M NaNO3 solution from a 15M NaNO3 solution, you can use the dilution formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution to be used
C2 = desired final concentration
V2 = final volume of the diluted solution
In this case:
C1 (concentration of the 15M NaNO3 solution) = 15M
C2 (desired final concentration) = 6.0M
V2 (final volume of the diluted solution) = 250 mL (0.250 L)
Now, plug in the values and solve for V1 (volume of the stock solution to be used):
15M * V1 = 6.0M * 0.250 L
V1 = (6.0M * 0.250 L) / 15M
V1 = 0.100 L
Thus, you would need to mix 0.100 liters (or 100 mL) of the 15M NaNO3 solution with enough solvent (e.g., water) to bring the total volume to 250 mL in order to obtain a 6.0M NaNO3 solution.
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M stands for molarity and V is volume. Plug in the information you have, and then solve for the unknown using simple algebra. You get 6.0M NaNo3 x 250mL= 15M NaNO3 x V2. Now solve for V2. Keep in mind that you will have to use stoichiometry to convert to liters instead of liters if your answer requires it. Hope this helps! :)
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Li₂S = 45.947 g/mol
AlCl₃ = 133.34 g/mol
3 Li₂S + 2 AlCl₃ = 6 LiCl + Al₂S₃
3 * 45.947 g Li₂S ----------> 2 * 133.34 g AlCl₃
1.084 g Li₂S ----------------> ?
Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947
Mass Li₂S = 289.08112 / 137.841
Mass Li₂S = 2.0972 g
hope this helps!