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Rushwr

  • one year ago

Fun but long ! @woodward and @photon336

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  1. Rushwr
    • one year ago
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    In an experiment to determine the Ksp of Ca(OH)2 a student followed the following procedure. 1) Dissolved Ca(OH)2 in 250cm^3 of water 2) After the solution was saturated , 3 solutions (25cm^3 volume in each) was separated out from the stock solution and phenolphthalein indicator was added to them. 3) Finally these 3 solutions were titrated with 0.05 moldm^-3 HCl solution and readings were taken. these were the reading of the 3 solutions. 12.5 12.05 and 11.95 all in cubic centimeters.

  2. Rushwr
    • one year ago
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    Question 1) By using the given data calculate the water solubility of Ca(OH)2 2) What is the colour change at the end point 3) Name another indicator that can be used for this titration 4) why is it necessary to do the titration 3 times 5) How can u identify that the solution is saturated with Ca(OH)2 ? 6) By using the above procedure is it possible to determine the Ksp of CaCO3 . Give reasons.

  3. Rushwr
    • one year ago
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    I will post the other part after ul finish this part ! :) @Photon336 @Woodward

  4. Rushwr
    • one year ago
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    I think photon just left !!! So are u gonna try this now or later? @Woodward

  5. anonymous
    • one year ago
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    "0.05 moldm^-3 HCl solution" did you mean to write "cm" or is this really "dm" ? :O

  6. Rushwr
    • one year ago
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    really in cubic decimeter,

  7. Rushwr
    • one year ago
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    1 tip !!! Since they have 3 burette readings for all 3 trials we have to take an average value for the volume of HCl so we have to add all 3 and divide it by 3 !!!

  8. Rushwr
    • one year ago
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    when u get the average round it off to the nearest whole number

  9. anonymous
    • one year ago
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    Ok I can answer most of these without finishing my calculations so I'll go for it, then finish up answering in a little while longer: 2) What is the colour change at the end point It should go from fuschia to colorless since it's going from basic (OH-) and then we're titrating it by adding acid. 3) Name another indicator that can be used for this titration Uhhh I don't know how indicators work exactly but I don't think it matters for strong acid/base titrations since they all have an equivalence point at pH=7. 4) why is it necessary to do the titration 3 times So that we can be more confident in our data! 5) How can u identify that the solution is saturated with Ca(OH)2 ? It precipitates to the bottom! 6) By using the above procedure is it possible to determine the Ksp of CaCO3 . Give reasons. I don't think so cause CO3- is a weak base so I think we might have to change things not sure.

  10. anonymous
    • one year ago
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    So first, \[K_{sp} = [Ca^{2+}] [OH^-]^2\] from the reaction equation, we know that the concentration of OH- in solution is twice that of Ca2+ so I'll just substitute this in: \[[Ca^{2+}]= \frac{1}{2}[OH^-]\] \[K_{sp} = \frac{ [OH^-]^3}{2}\] Now I just need to determine the concentration of OH- from titration! To make it easier for myself, I just do this part of the division first: \[M_1 V_1 = M_2 V_2\] \[V_1 \frac{M_1}{V_2} = M_2\] \[\frac{M_1}{V_2} = \frac{.05 \frac{mol}{dm^3}}{26 \ cm^3} = 2 \times 10^{-6}\frac{mol}{cm^6} \] So I take all the volumes and multiply them by this factor to get the molar concentration of [OH-] \[12.5 \ cm^3 \times 2 \times 10^{-6}\frac{mol}{cm^6} = 2.5 \times 10^{-5} \frac{mol}{cm^3}\] \[12.05 \ cm^3 \times 2 \times 10^{-6}\frac{mol}{cm^6} = 2.41 \times 10^{-5} \frac{mol}{cm^3}\] \[11.95 \ cm^3 \times 2 \times 10^{-6}\frac{mol}{cm^6} = 2.39 \times 10^{-5} \frac{mol}{cm^3}\] Now I average these and plug it into my formula for Ksp: \[K_{sp} = \frac{(2.43 \times 10^-5)^3}{2}=3.6 \times 10^{-15}\] I found that the real data online for it is \(5.5×10^{–6}\) So I think something went wrong along the way.

  11. Rushwr
    • one year ago
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    11.95 , 12.05 and 12.5 are burette readings of HCl , we rounded of the average value and got 12 right? Use that value everywhere. And the answer u have gotten is wrong

  12. Rushwr
    • one year ago
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    I'll put the answer after photon replies @Photon336 we are waiting for u !

  13. anonymous
    • one year ago
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    even if I use 12, that's not enough to fix it. That volume of acid is when the moles of acid is equal to the moles of OH- isn't it? And that's how I'm finding the concentration of CaOH2

  14. Rushwr
    • one year ago
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    We can't relate concentration with the stoichiometric number, we only can compare the moles. I think u should go by tht

  15. Rushwr
    • one year ago
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    Hey why don't ul try this too ?? @sweetburger @arindameducationusc

  16. anonymous
    • one year ago
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    \[Ca(OH)_2 \rightarrow Ca^{2+}+2 OH^-\] Why is this not true: \[2[Ca^{2+}] = [OH^-]\] I think it's right.

  17. Photon336
    • one year ago
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    Sorry guys my OS crashes

  18. Rushwr
    • one year ago
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    huh

  19. Rushwr
    • one year ago
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    U wanna try @RamiroCruzo

  20. RamiroCruzo
    • one year ago
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    Well always ready for challenges.......... :D

  21. RamiroCruzo
    • one year ago
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    (1)For the question, \[Ca(OH)_{2}\rightarrow Ca ^{+2} + 2OH ^{-}\] s s 2s Now, \[25\times2s=0.05\times12.25\] \[s=1.225\times10^{-2}\] \[K _{sp}=s.(2s)^{2}=4s ^{3}=4(1.225\times10^{-2})^{3}=7.35\times10^{-6}\]

  22. RamiroCruzo
    • one year ago
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    @Rushwr Am I right Ma'am ;)????

  23. Rushwr
    • one year ago
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    NOP I'll post the answer 2mrw !!! :)

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