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As chemistry titration help

Hi guys! I'm stuck on this titration calculation . All help is appreciated.

The burrette contains 0.1 mol dm-3 NaOH.
The acid is Harpic (a powder) which is soluble in water so can be made into a solution. The active component of Harpic is a acid salt of forumula MHS04 (where M is a group 1 metal). in solution it behaves as a monobasic acid as it ionised completely to give h+, m+, so4 2- .

3.337 g of harpic was dissolved in 50cm3 of water in a beaker. This was added to a 250cm3 volumetric flask, and water was added until the solution was 250cm3. Using a pipette, 25cm3 of this solution was placed in conical flask.

2 drops of phenalphlation indication was added and NaOH solution was also added until the end point was reached.

The average titre , i.e amount of NaOH required to neutralise 25cm3 of the acid solution was 22.2 cm3.

QUESTION

1) from the titre and weightings, calculate the mass of harpic which contains exactly one mole of h+ ions, i.e. one mole of MHSO4. Hence suggest the identity of M.

(We were told there is 1:1 reaction ratio, so Moles of NaOH = (cxv)/1000=0.00222. this means there are 0.00222 moles of MHS04 as well. However i am completely stuck after this point)

Please note ALL help is much appreciated. Thank you

Reply 1

Nitrogen

(22.2/1000)*0.1 = n of NaOH= 0.00222
Its a neutralisation reaction so harpic has the same moles too.
multiply moles by 10 to get how many moles in the 250cm3 solution
mass/mr =n so 3.337/mr =0.0222

Reply 2

Pigster

Your calculation is correct.

I have a couple of niggles with some of the English used, e.g. "i.e amount of NaOH required to neutralise 25cm3", i.e. you meant volume : )

Out of interest, if you got your 25cm3 and your 22.2cm3 the wrong way around, the you'd get 21.4 at the atomic mass, which is fairly close to Na, but still not close enough for my liking.

Reply 3

cmgreenland

Hi there,
So you have worked out that the number of moles of MHSO4 in the 25cm3 titrated however there is 10 x that number of moles in the volumetric flask, so the number of moles in the volumetric flask is 10 x 0.00222 = 0.0222 moles of MHSO4. So therefore, 3.337g of harpic powder produces 0.0222 moles of MHSO4 and we then find the molar mass of the MHSO4 by using the equation Number of Moles = Mass/Molar mass. The molar mass of the harpic powder is therefore 3.337/0.0222 = 150.3 and the mass of harpic powder containing mole mole of H+ ions would therefore be 150.3g . To work out what M was, we would have to subtract the atomic masses of HSO4 from the molar mass of the MHSO4 to find the atomic mass of M. 150.3 -32 -16-16-16-16-1 = 53.3 atomic mass which most likely Potassium if M is a group 1 metal however it is very far away from Potassium's atomic mass of 39. Are you sure you copied down the numbers correctly ?