The following picture represents the hydrated metal cation M1H2O26 n + , where n = 1, 2, or 3.

(b) Does the equilibrium constant increase, decrease, or remain the same as the value of n increases? Explain.

Video transcript

Hi everyone for this problem. It reads an acquis solutions. The acidity ease of methanol and water are almost identical and are increased when coordinated with a central metal cat eye on the resulting metal complex then acts as an acid and donates a hydrogen ion to a base shown below is a metal cat ion with methanol Liggins attached to it. What happens to the equilibrium constant? Does it increase, decrease or remain the same as the value of n increases, explain. Okay, so looking at what we have, the the molecule given is going to act as an acid. Okay. And it's going to act as an acid since the methanol ligand electrons are pulled towards the positive central atom and this is going to leave the hydroxide bonds more polarized than usual and the hydrogen nuclear exposed for attack. Okay, so in a quick solutions, the metal complex donates a hydrogen ion to water molecules in the solution. And I'm going to write the equilibrium reaction for this. Okay. And that is going to be okay. So this is the equilibrium reaction for this. And looking at the image shown below, when the lone pairs on the oxygen atom of methanol form coordinate bonds with the central metal ion electrons move towards the metal. Okay, so what this is going to cause is the electrons in the hydroxide bond are pulled towards the oxy even more than usual, leaving the hydrogen nuclear more exposed and each of the hydrogen atoms become more positive than it usually is an ordinary methanol molecules. So the hydrogen atoms attached to the methanol Liggins are sufficiently positive such that they pull off in a reaction involving water molecules in the solution. So to answer the question of what happens to the equilibrium constant as the value of an increases in higher charged central atoms, the electrons in the hydroxide bones are pulled farther away from the terminal hydrogen than in lower charged ions. As the value of N increases, the equilibrium constant is also going to increase since the hydroxide bonds from the attached methanol molecules become more polarized. So the answer that we want to look for here is the answer that reflects the value of the equilibrium constant, increasing. Okay, and when we take a look at our answer choices, we see that C is going to be our answer choice. The value of the equilibrium constant increases since the hydroxide bonds become more polar and the hydrogen atom dissociation is more favorable. That is it for this problem, I hope this was helpful.