# Acid-based titration calculationWatch

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#1
I'm stuck on this practice question for A-level Chemistry:

An indigestion tablet containing sodium hydrogencarbonate was dissolved in 250cm^3 of water. 25.0cm^3 of this solution was found to need 16.50cm^3 of a 0.150mol dm^-3 solution of HCl for neutralisation.

NaHCO3+HCl --> NaCl + H2O + CO2

If the original mass of the indigestion tablet was 3.20g, calculate the percentage by mass of sodium hydrogencarbonate in the tablet.

I've seen someone say the correct answer is 32.5% but I keep getting 26% for some reason. Could someone show me how to work it out?
Any help would be much appreciated.
0
1 year ago
#2
(Original post by AndiP4ndi)
I'm stuck on this practice question for A-level Chemistry:

An indigestion tablet containing sodium hydrogencarbonate was dissolved in 250cm^3 of water. 25.0cm^3 of this solution was found to need 16.50cm^3 of a 0.150mol dm^-3 solution of HCl for neutralisation.

NaHCO3+2HCl --> NaCl + H2O + CO2

If the original mass of the indigestion tablet was 3.20g, calculate the percentage by mass of sodium hydrogencarbonate in the tablet.

I know the correct answer is 32.5% but I keep getting 26% for some reason. Could someone show me how to work it out?
Any help would be much appreciated.
show us how you get 26% and we'll point out your error ...
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#3
(Original post by charco)
show us how you get 26% and we'll point out your error ...
This is my working:

.........NaHCO3 + HCl
Conc................0.150
Vol..................0.0165
Moles 0.002475....0.002475
Ratio 1:1

16.50cm^3 / 1000 = 0.0165 dm^3
0.150 x 0.0165 - 0.002475 moles

.......NaHCO3
Mass 0.2079
Mr 84
Moles 0.002475

Mr = 23 + 1 + 12 + 3(16) = 84
0.002475 x 84 = 0.2079g

0.2079g / 0.250 = 0.8316

(0.8316/3.2) x 100 = 25.9875%
= 26.0% (3 sf)

I've never done this type of calculation before so I don't know where I've gone wrong. Thanks for any help.
0
1 year ago
#4
(Original post by AndiP4ndi)
I'm stuck on this practice question for A-level Chemistry:

An indigestion tablet containing sodium hydrogencarbonate was dissolved in 250cm^3 of water. 25.0cm^3 of this solution was found to need 16.50cm^3 of a 0.150mol dm^-3 solution of HCl for neutralisation.

NaHCO3+2HCl --> NaCl + H2O + CO2

If the original mass of the indigestion tablet was 3.20g, calculate the percentage by mass of sodium hydrogencarbonate in the tablet.

I know the correct answer is 32.5% but I keep getting 26% for some reason. Could someone show me how to work it out?
Any help would be much appreciated.
Here is the worked solution which gives you % mass of 32.5%.

1. n(H+) titration = (16.5/1000) x 0.15 = 0.002475mol
work out the moles of H+/HCl needed to neutralise the HCO3- ion in the titration.

2. n(HCO3-) = 0.002475 / 2 = 0.0012375mol
look at the mole ratio in the reaction you provided. For every one HCO3- ion there were 2 H+ ions reacting. So we divide the moles of H+ by 2 to get moles of HCO3- in the titration.

3. n(HCO3-) total = 0.0012375 x 10 = 0.012375mol
remember that the mol of HCO3- calculated in step 2 was only in 25cm3 solution. Originally the tablet was dissolved in 250cm3, which is 10 times the volume. So the total amount of HCO3- is 10 times the amount in the titration. (i keep mentioning HCO3- because the NaHCO3 is ionic and will dissociate when in the solution)

4. mass of NaHCO3 = 0.012375 x (23 + 1 + 12 + 48) = 1.0395 g
now we know moles of NaHCO3 ( moles of HCO3- = moles of Na+ in the ionic lattice) and its Mr, we work out its mass

5. % mass = (1.0395 / 3.2) x 100 = 32.484375% which rounds to 32.5%
0
1 year ago
#5
An indigestion tablet containing sodium hydrogencarbonate was dissolved in 250cm^3 of water. 25.0cm^3 of this solution was found to need 16.50cm^3 of a 0.150mol dm^-3 solution of HCl for neutralisation.

NaHCO3+HCl --> NaCl + H2O + CO2

If the original mass of the indigestion tablet was 3.20g, calculate the percentage by mass of sodium hydrogencarbonate in the tablet.

(Original post by AndiP4ndi)
This is my working:

.........NaHCO3 + HCl
Conc................0.150
Vol..................0.0165
Moles 0.002475....0.002475

Ratio 1:1
up to this point you are OK, but you now need to multiply by 10 to get moles equivalent in 250ml

= 0.02475 mol

You then find out the mass of this number of moles of sodium hydrogencarbonate

= 0.02475 x 84 = 2.079g

and now use the actual mass to find the percentage purity

= 100 * 2.079/3.02

=68.9%

16.50cm^3 / 1000 = 0.0165 dm^3
0.150 x 0.0165 - 0.002475 moles

.......NaHCO3
Mass 0.2079
Mr 84
Moles 0.002475

Mr = 23 + 1 + 12 + 3(16) = 84
0.002475 x 84 = 0.2079g

0.2079g / 0.250 = 0.8316

(0.8316/3.2) x 100 = 25.9875%
= 26.0% (3 sf)

I've never done this type of calculation before so I don't know where I've gone wrong. Thanks for any help.
0
1 year ago
#6
(Original post by dip0)
Here is the worked solution which gives you % mass of 32.5%.

1. n(H+) titration = (16.5/1000) x 0.15 = 0.002475mol
work out the moles of H+/HCl needed to neutralise the HCO3- ion in the titration.

2. n(HCO3-) = 0.002475 / 2 = 0.0012375mol
look at the mole ratio in the reaction you provided. For every one HCO3- ion there were 2 H+ ions reacting. So we divide the moles of H+ by 2 to get moles of HCO3- in the titration.

3. n(HCO3-) total = 0.0012375 x 10 = 0.012375mol
remember that the mol of HCO3- calculated in step 2 was only in 25cm3 solution. Originally the tablet was dissolved in 250cm3, which is 10 times the volume. So the total amount of HCO3- is 10 times the amount in the titration. (i keep mentioning HCO3- because the NaHCO3 is ionic and will dissociate when in the solution)

4. mass of NaHCO3 = 0.012375 x (23 + 1 + 12 + 48) = 1.0395 g
now we know moles of NaHCO3 ( moles of HCO3- = moles of Na+ in the ionic lattice) and its Mr, we work out its mass

5. % mass = (1.0395 / 3.2) x 100 = 32.484375% which rounds to 32.5%
The ratio is 1:1

NaHCO3 + HCl --> NaCl + H2O+CO2
1
#7
Thank you! I got 64.9% when I tried it again (I think you meant 64.9 and not 68.9). Thanks for the help.
0
#8
(Original post by charco)
up to this point you are OK, but you now need to multiply by 10 to get moles equivalent in 250ml

= 0.02475 mol

You then find out the mass of this number of moles of sodium hydrogencarbonate

= 0.02475 x 84 = 2.079g

and now use the actual mass to find the percentage purity

= 100 * 2.079/3.02

=68.9%
Thank you! I got 64.9% when I tried it again (I think you meant 64.9 and not 68.9). Thanks for the help.
0
#9
..
0
1 year ago
#10
(Original post by charco)
The ratio is 1:1

NaHCO3 + HCl --> NaCl + H2O+CO2
Oops my bad! Yes that's true. Thanks for the correction.
0
1 year ago
#11
(Original post by AndiP4ndi)
Thank you! I got 64.9% when I tried it again (I think you meant 64.9 and not 68.9). Thanks for the help.
actually I meant 68.8%
0
#12
(Original post by charco)
actually I meant 68.8%
But 100 * 2.079/3.2 = 64.96875 (which would round up to 65)
0
1 year ago
#13
(Original post by AndiP4ndi)
But 100 * 2.079/3.02 = 64.96875 (which would round up to 65)
Not on my calculator it's not ...

0
1 year ago
#14
(Original post by AndiP4ndi)
But 100 * 2.079/3.02 = 64.96875 (which would round up to 65)
the mass of tablet is 3.2g, not 3.02 grams.
Thats why you are getting 64.9...% instead of 68.8%
0
1 year ago
#15
(Original post by charco)
Not on my calculator it's not ...

The mass of the tablet is 3.2g, not 3.02g.
So the % mass is 2.079/3.2 x 100 = 64.9...%

You got 68.8...% because you did 2.079/3.02 x 100
0
1 year ago
#16
(Original post by dip0)
The mass of the tablet is 3.2g, not 3.02g.
So the % mass is 2.079/3.2 x 100 = 64.9...%

You got 68.8...% because you did 2.079/3.02 x 100
very true ...
0
4 weeks ago
#17
(Original post by charco)
The ratio is 1:1

NaHCO3 + HCl --> NaCl + H2O+CO2
sorry im commenting late but its 2HCl so the ratio isnt 1:1 but 2mol hcL:1mol NaHCO3 so the answer is right
0
4 weeks ago
#18
(Original post by Hassara otsusuki)
sorry im commenting late but its 2HCl so the ratio isnt 1:1 but 2mol hcL:1mol NaHCO3 so the answer is right
The ratio is 1:1.
0
4 weeks ago
#19
The ratio is 1:1.
what do you mean
0
4 weeks ago
#20
(Original post by Hassara otsusuki)
what do you mean
oh my one has 2HCl
0
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