# pH calculation????

#1
0.5g of a mixture of sodium hydrogencarbonate and anhydrous sodium carbonate was dissolved in water. The solution was titrated against 0.1M hydrochloric acid, requiring 15cm^3 using phenolphthalein as indicator. Calculate the percentage by mass of sodium hydrogencarbonate in the mixture.

I have no idea where to start, ,i don't even know what to do, someone set me in the right direction.
0
5 years ago
#2
Sodium carbonate has two end points, one of them is the same as the end point for the hydrogencarbonate end point.

Which one does the phenolphthalein indicator change colour on?
0
#3
(Original post by Pigster)
Sodium carbonate has two end points, one of them is the same as the end point for the hydrogencarbonate end point.

Which one does the phenolphthalein indicator change colour on?
It does? This is new to me.....

I haven't done anything redox.

I recently found this was a redox titration which i know nothing about, how nice of my teacher to put a redox titration question on a pH calculations sheet
Well so far i found some stuff on past papers and some similar questions that's how i got 48g for that other question
0
#4
(Original post by Pigster)
Sodium carbonate has two end points, one of them is the same as the end point for the hydrogencarbonate end point.

Which one does the phenolphthalein indicator change colour on?
But it's ok

NaHCO3+Na2CO3+3HCl--->3NACl+2H2O+2CO2

found mol of HCl=1.5x10^-3
3 mol of HCl=1 mol of NaHCO3

didve by 3

mol of NaHCO3=5x10^-4

mass=molxMr
=5x10^-4 x 84=0.042g

0.042/0.5 all multiplied by 100 gives me 8.4%
0
5 years ago
#5
It isn't a redox titration.

The phenolphthalein changes colour when all of the CO32- converts into HCO3-.

n(HCl) required to reach this endpoint = c v = 0.1 x 15/1000 = 0.0015 mol = n(Na2CO3)

m(Na2CO3) = n x Mr = 0.0015 x 106 = 0.159 g

The rest is NaHCO3, i.e. 0.341 = 68.2%

But I am sleepy.
0
#6
(Original post by Pigster)
It isn't a redox titration.

The phenolphthalein changes colour when all of the CO32- converts into HCO3-.

n(HCl) required to reach this endpoint = c v = 0.1 x 15/1000 = 0.0015 mol = n(Na2CO3)

m(Na2CO3) = n x Mr = 0.0015 x 106 = 0.159 g

The rest is NaHCO3, i.e. 0.341 = 68.2%

But I am sleepy.
I see thank you for your help. Tbh tho this hwk i spent way too long on and it's in for tomorrow anyway so not really much point replying besides my piece of mine
But surely i need a balanced equation for this? Or how would i know it's a 1:1 ratio?
0
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