"Homeworkify by Tutor AI: Your AI-powered learning companion"
I was interested to know if this would give you answers and explanations, like ChatGPT etc (which are notorious for being used to do homework), or whether instead of just directly giving you answers and explanations, it would try to guide you to figuring out the answer on your own, the way a good human tutor would. To find out, I figured I would need an actual homework problem, so I cracked open my chemistry book and grabbed a problem:
An aqueous solution has an iodine concentration of 2.00 * 10^-3 mol/L. Calculate the percentage of iodine remaining in the aqueous phase after extraction of 0.100 L of this aqueous solution with 0.050 L of CCl4 at 25 degrees C. The equilibrium constant K for [I2]CCl4 / [I2]aq = 85 at 25 degrees C.
Here's the reply:
Let's break down this problem step by step to find the percentage of iodine remaining in the aqueous phase after the extraction.
Step 1: Define the Equilibrium Constant
The equilibrium constant K$$ for the partitioning of iodine between carbon tetrachloride $ CCl4 $ and water is given by the expression: $$ K = \frac{$I_2${CCl4}}{I2{aq}} = 85 HereI_2${CCl4} istheconcentrationofiodineintheCCl4phaseandI2{aq}$ is the concentration of iodine in the aqueous phase.
Step 2: Initial Concentrations
The initial concentration of iodine in the aqueous solution is:
$$ $I_2${aq \text{initial}} = 2.00 \times 10^{-3} \text{ mol/L}
Theinitialvolumeoftheaqueoussolutionis: V{aq} = 0.100 \text{ L}
ThevolumeofCCl4usedforextractionis: V_{CCl4} = 0.050 \text{ L} $$
Step 3: Set Up the Equilibrium Expressions
Let x be the amount of iodine inmoles that transfers from the aqueous phase to the CCl4 phase.
The iodine concentration in the aqueous phase after extraction is:
$$ $I_2${aq \text{final}} = \frac{ 2.00 x 10^-3 mol/L \cdot 0.100 \text{ L} - x}{0.100 \text{ L}}
TheiodineconcentrationintheCCl4phasewillbe: I2{CCl4} = \frac{x}{0.050 \text{ L}} $### Step 4: Apply the Equilibrium Constant
Using the equilibrium constant expression:$ K = \frac{$I_2${CCl4}}{I2{aq \text{final}}} = 85
SubstitutetheexpressionsforI_2${CCl4} and I2{aq \text{final}}$:
85 = \frac{\frac{x}{0.050}}{\frac{$ 2.00 \times 10^{-3} \times 0.100 $ - x}{0.100}}
Step 5: Solve for x
Simplify the equation:
85 = (x / 0.050) / (((2.00 x 10^-4) - x) / 0.100)
85 = 2x / (0.2 x 10^-3 - x)
Now solve for x:
85 $ 0.2 \times 10^{-3} - x $ = 2x 17 \times 10^{-3} - 85x = 2x
17 x 10^-3 = 87x
x = \frac{17 \times 10^{-3}}{87} $$ x \approx 1.95 \times 10^{-4} \text{ mol} $### Step 6: Calculate the Final Concentration in Aqueous Phase
The final concentration of iodine in the aqueous phase is: I_2${aq \text{final}} = \frac{2.00 x 10^-3 x 0.100- 1.95 \times 10^{-4}}{0.100} $$ $I_2${aq \text{final}} = \frac{2.00 \times 10^{-4} - 1.95 \times 10^{-4}}{0.100} $$ $I_2${aq \text{final}} = \frac{0.05 \times 10^{-4}}{0.100} $$ $I_2${aq \text{final}} = 0.50 \times 10^{-3} \text{ mol/L} $### Step 7: Calculate the Percentage of Iodine Remaining
The percentage of iodine remaining in the aqueous phase is: \text{Percentage remaining} = \left \frac{I2{aq \text{final}}}{I2{aq \text{initial}}} \right x 100 \text{Percentage remaining} = \left \frac{0.50 \times 10^{-3}}{2.00 \times 10^{-3}} \right x 100 \text{Percentage remaining} = \left \frac{0.50}{2.00} \right x 100 Percentage remaining = 25$
Conclusion
After the extraction 25% of the iodine remains in the aqueous phase.
Yes, it really did give me a malformatted mishmash of markdown and LaTeX, with words mashed together and line breaks in the wrong places and the whole deal. So, that's one strike against it right there. I will have to try the AI LaTeX helper next to see if I can make this all nicely formatted.
Secondly, it definitely didn't do any interactive back-and-fourth like a human tutor would to guide me into figuring out the answer on my own and learning the key concepts in the process. So I definitely see an opening for an AI tutor that can do that.
Thirdly, it got the answer wrong. It said 25% but the correct answer is 2.3%.
However, it did have the right general idea. You make a variable x that remains in the aqueous phase -- or you could do the reverse and define x as the amount that changes to the CCl4 phase, which is what it actually decided to do. 2.00 x 10^-4 - x will be what's in the other phase, regardless of which way you define it.
So, it set the solution up correctly but somewhere in the plugging and chugging it did the algebra wrong.
