Consider an amphoteric hydroxide, M(OH)2(s), where M is a...

Consider an amphoteric hydroxide, M(OH)2(s), where M is a generic metal. M(OH)2(s) ? M2+(aq) + 2OH-(aq) Ksp = 9 × 10^-16 M(OH)2(s) + 2OH-(aq) ? [M(OH)4]2-(aq) Kf = 0.08 Estimate the solubility of M(OH)2 in a solution buffered at pH = 7.0, 10.0, and 14.0. solubility at pH = 7.0 solubility at pH = 10.0 solubility at pH = 14.0

Transcript

00:01 Hello, so here the question is based on solubility.

00:04 Okay, so here in this question we are given with an amphoteric hydroxide.

00:10 Okay, amphoteric hydroxide that is m -o -h twice, okay, where m is a general matter.

00:20 Now, here we are given with two equations, that is m -o -h twice solid is dissociating to give m -2 positive plus 2 o h minus here the ksp value is given as 9 multiplied by 10 to the power negative 60 okay then the next equation is m o h twice plus 2 o h minus is in equilibrium with m o h 4 times 2 minus okay so here the kf value is given as 0 .08 okay now we need to find the solubility our ph is equal to 7, 10 and 14.

01:04 Okay, so let's see the answer.

01:05 First we can consider ph is equal to 7.

01:08 So when ph is equal to 7, we know the equation ph plus p -o -h is equal to 14.

01:15 So from here we will get p -o -h is equal to 14 minus 7 and that is equal to 7.

01:22 So from here the concentration of o -h minus will be equal to 1 multiplied by 10 to the power negative 7.

01:28 Okay then what we need to calculate here is for the concentration of m2 positive so we have the equation ksp is equal to concentration of m2 positive multiplied by concentration of o h minus the whole square so here we are considering this equation okay so we can substitute to calculate so that is nine multiplied by 10 to the power negative 16 equal to concentration of m2 positive multiplied by 10 to the power negative 7 squared okay so this will give the answer as concentration of m2 positive is equal to 9 multiplied by 10 to the power negative 16 divided by 10 to the power negative 14 so that is equal to 9 multiplied by 10 to the power negative to molar concentration okay so we can write the solubility at ph 7 is equal to 9 multiplied by 10 to the power negative to molar concentration similarly we can calculate at ph is equal to 10.

02:30 Okay, so when ph is equal to 10, p -o -h will be equal to 14 minus 10 and that is equal to 4.

02:37 So this will give the concentration of o -h minus ions as 10 to the power negative 4.

02:42 Okay.

02:43 So we can substitute in ksp.

02:46 So k -sp is equal to concentration of m -2 positive ions multiplied by the concentration of o -h minus the square...

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