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What is the molar solubility of $\mathrm{T} ](\mathrm{OH})_{3}$ in a $0,10-\mathrm{M}$ solution of $\mathrm{NH}_{3} ?$
$2.68 \times 10^{-37} \mathrm{M}$
Chemistry 102
Chapter 15
Equilibria of Other Reaction Classes
Chemical Equilibrium
University of Central Florida
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let me start by writing the equilibrium reaction Richards. And that's three less. It's tool, which gives out an itch. Four. Positive less which make it. Do. We already know the base dissociation. Constant expression. So let x b the concentration off which negative I informed. When equilibrium is reached, we substitute X for the next full or the depth So Ex and X, which naked, divine. For the next three, it will be 0.10 minus X and be already provided with Gaby value. So substituting all this and baby formula, we have 1.8. But to play 10 days to power negative five for K B equals ex multiplied with ex form product divided by the 0.1 Tzeitel minus x So taking everything towards the one side off equation the half X square less 1.8 envies to one negative. Five x minus 0.1 Tzeitel 1.8 and a stupid negative for you. He couldn't see, you know, solving for X. We have to a 0.66 10 days to one negative three do I did by two which don't know 1.33 Send these too negative three more So we know Excellent presents The concentration off which negative ions. So also which negative and concentration we substitute this 1.33 value. What? Which? Negative Ryan into GSP formula. We know. Okay, SP you see? Well, Toe buddy reported To which negative based Power three and Gatsby is 6.3. My reply. 10 days to burn. Negative 46 equals Dean concentration with 1.33 What? The Pretenders to one negative three. Nice to poetry. So the consul. Sorry. The moonless vulnerability. Oh e A Which three? Donato be 2.68 Multiply 10 days to one negative 37 morning.
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