Answers by visitors named: Aanya

x^2 +2ax+b is the expression #PsyDAG

Thank you very much Emily! And if we titrate 25cm^3 of the final solution with 0.1 M HCl,using Me.O as the indicator,is there any reaction taking place? I think no as they are both acids(KOH is reacted completely) and the concentrations of OH- we get from hydrolysis of water is a very low value (1*10^-7 M) and also the Concentrations of other ions such as KHS X2- are too small to react with HCl of 0.1 M What do you think?

Let the shortest side be x(cm).So the longest side equals to (x+10)cm and the other side equals to (x+x+10)/2 So perimeter of the triangle= x + (x+10)+[(x+x+10)/2] =60 2x+10+[(2x+10)/2] =60 Now let's multiply both sides by 2 to get, 2(2x+10) + (2x+10) = 60*2 4x+20 +2x+10 = 129 6x+30=120 6x=90 x= 15 cm

There's a typo in the in the fourth line from the bottom.It should be corrected as, 4x +20 +2x +10 = * 120 * is to indicate the place I've corrected

Thank you very much DrBob222.I totally forgot about the Henderson equation! I think you cleared out all my doubts.Thank you very much for the help..

Thank you very much for your concern on this question in the first place. And I forgot to include that It is given that the temperature is 25°C. Again I'm also having doubts.I will wait for your answer and in the mean time I'll try this again.

Thank you very much! Now it sounds more reasonable! And I have another question regarding this. If we titrate the final solution with NaOH of same concentration what are the changes which should be done in this experiment?

Thank you very much! Now it sounds more reasonable! And I have another question regarding this. If we titrate the final solution with NaOH of same concentration as HClmentioned in the last part what are the changes which should be done in this experiment?

DrBob222 don't we have to consider the hydrolysis of HX- here? H2O(l)+ HX-(aq)<==>H2X(aq)+OH-(aq)-->(1) equ. (0.1-x) x x (moldm-3) Kh= [H2X][OH-]/[HX-] Kh=kw/ka(HX-) But we are not given the kW value and we can take it as 14 only if the temperature is 25°C. Let's assume the temperature is 25°C and kW=14,which gives kh= 10^(-14)/[5.3*(10)^-9]= (1/5.3)*10^(-5) moldm-3=[x^2/(0.1-x)] if we take (0.1-x) is approximately 0.1,we get, x^2 = 0.19*(10)^-5==> 2*(10)^-5 M x==> 4.46*(10)^-2 But we've been given the ka of H2X. H2X<===> H+(aq) + HX-(aq)->(2) equ. (x-y) y y (moldm-3) which gives, y^2 =9*(10)^-5 moldm-3 * x y^2 =9*4.46*(10)^-7 M Don't we have to consider both (1) and (2) like this to find pH

Some corrections should be added! x^2==>2*(10)^-6 M x==>1.41*(10)^-3 M y^2= 9*1.41*(10)^-8 M y^2=12.69*(10)^-8 M y ==>3.56*(10)^-4 M [H+]=> 3.56*(10)^-4 M

Emily there are two reactions of H2X with NaOH ,as of what I've learned.. H2X+NaOH ---> NaHX + H2O when there's excess NaOH, H2X+ 2NaOH ---> Na2X +2H2O

First find the moles of H2X with the help of the NaOH moles.Now you get the moles of H2X in 25ml.Multiply it by 4 to find the moles in 100ml. Find the concentration of NaOH with the given data in the titration of KHP. Find the molar mass of H2X by dividing the no.of H2X moles by the given mass of H2X n=m/M) n=no.of moles m=given mass M=molar mass

So what are the changes which should be made,if we titrate the final solution with NaOH,not with HCl? Change of indicators or molarities etc.

Sn^(2+)?

Thank you! I was worried about ths question stating Fe3+ becoming colourless

Yes that is what I meant! :-)

*H2CO will create?

Is that what you meant?

1 picometer= 1*(10)^-12 m

Than O2?

The same here :-(

Thanks a lot!!

Radius of the beaker?

*1/2m(w^2)=aRd

But they have given it like that in the question, that A reaches the previous point.

E°cell= E°cathode-E°anode E°cell=-0.44-(-0.74) = 0.30V Ecell= E°cell - 0.0591/n log [Cr3+]^2/[Fe2+]^3 Ecell= 0.30- 0.0591/6 log[0.1]^2/[0.1]^3 Ecell= 0.30 - 0.0591/6 log 10 Ecell= 0.30 - 0.0591= 0.2409V