2 Questions :) (1 Viewer)

[wrxsti]

1) 2Na + 2h20 --> 2NaoH + H2
pure sodium metal dropped into 1.2L of water...
Gas collected occuped a volume of 4.68L at 25degrees caculate the final PH of the water NOTE: this is a 2002 paper so u have to use 1mole at 25degrees is 24.47L


2) In a titration the electrical conductivity of the reaction starts at a maximum then decreases to a minimum why does it NOT reach zero

3) Sorry one more.... how do you ensure Anhydroud NaCO3 remains anhydrous
THANKSSS IN ADVANCE

 

[TesseracT22]

ok then,

1) amount of H2=4.68L = 4.68/24.47 moles = 0.19 moles of H2
therefore 0.38 moles of OH- have been added into the solution from loss of H+.

originally in water there is 1*10^-7 molL^-1 concentration of OH- and so
in 1.2 L there was 1.2*10^-7 moles of OH-.

therefore now there are 1.2*10^-7 + 0.38 moles of )H- = 0.38 moles OH-
in 1.2 L
and so [OH-] = 0.32 molL^-1.

[H+][OH-] = 10^-7 and so [H+] = 3.16*10^-7molL^-1 therefore pH = -log[H+]

pH= -log[3.16*10^-7] = 6.5


2)When you titrate acids and bases you usually have a salt being formed, if this salt is soluble then it contributes to conductance of the solution aswell as the H+ and OH- ions so conductivity cant reach zero due to the presence of the salt.

3) not really sure about this one but maybe you could put it in a dessicator or use a dehydrating agent such as conc. H2SO4 to get rid of any hydration.

hope that helped

 

[xiao1985]

re: [H+][OH-] = 10^-7

you mean 10^-14?

pH should be: 13.50 (you have NaOH in water... the pH shouldn't be less than 7)

2) concur about the salt part. the self ionisation of water, which produces H+ and OH- has very negligible effect on conductivity however. Therefore, in titration of Ba(OH)2 (is this even soluable? oO) and H2SO4 for eg, you will notice conductivity drop to zero.

3) dessicator as teeserac said. or if it is wet, shove it in oven for couple of days.

 

[wrxsti]

thanxxxx