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Rushwr
 one year ago
Fun but long ! @woodward and @photon336
Rushwr
 one year ago
Fun but long ! @woodward and @photon336

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Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1In 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.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Question 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.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1I will post the other part after ul finish this part ! :) @Photon336 @Woodward

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1I think photon just left !!! So are u gonna try this now or later? @Woodward

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"0.05 moldm^3 HCl solution" did you mean to write "cm" or is this really "dm" ? :O

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1really in cubic decimeter,

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.11 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 !!!

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1when u get the average round it off to the nearest whole number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.111.95 , 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

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1I'll put the answer after photon replies @Photon336 we are waiting for u !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0even 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

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1We can't relate concentration with the stoichiometric number, we only can compare the moles. I think u should go by tht

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Hey why don't ul try this too ?? @sweetburger @arindameducationusc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[Ca(OH)_2 \rightarrow Ca^{2+}+2 OH^\] Why is this not true: \[2[Ca^{2+}] = [OH^]\] I think it's right.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Sorry guys my OS crashes

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1U wanna try @RamiroCruzo

RamiroCruzo
 one year ago
Best ResponseYou've already chosen the best response.0Well always ready for challenges.......... :D

RamiroCruzo
 one year ago
Best ResponseYou've already chosen the best response.0(1)For the question, \[Ca(OH)_{2}\rightarrow Ca ^{+2} + 2OH ^{}\] s s 2s Now, \[25\times2s=0.05\times12.25\] \[s=1.225\times10^{2}\] \[K _{sp}=s.(2s)^{2}=4s ^{3}=4(1.225\times10^{2})^{3}=7.35\times10^{6}\]

RamiroCruzo
 one year ago
Best ResponseYou've already chosen the best response.0@Rushwr Am I right Ma'am ;)????

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1NOP I'll post the answer 2mrw !!! :)
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