# Acid-based titration calculation Watch

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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.

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.

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#2

(Original post by

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.

**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.

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(Original post by

show us how you get 26% and we'll point out your error ...

**charco**)show us how you get 26% and we'll point out your error ...

.........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.

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#4

**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.

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%**
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#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.

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

This is my working:

.........NaHCO3 + HCl

Conc................0.150

Vol..................0.0165

Moles 0.002475....0.002475

Ratio 1:1

**AndiP4ndi**)This is my working:

.........NaHCO3 + HCl

Conc................0.150

Vol..................0.0165

Moles 0.002475....0.002475

Ratio 1:1

= 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.

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#6

(Original post by

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.

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%

**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%**NaHCO

_{3}+ HCl --> NaCl + H

_{2}O+CO

_{2}

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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.

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(Original post by

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%

**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%

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#11

(Original post by

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.

**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.

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(Original post by

actually I meant 68.8%

**charco**)actually I meant 68.8%

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#13

(Original post by

But 100 * 2.079/3.02 = 64.96875 (which would round up to 65)

**AndiP4ndi**)But 100 * 2.079/3.02 = 64.96875 (which would round up to 65)

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#14

(Original post by

But 100 * 2.079/3.02 = 64.96875 (which would round up to 65)

**AndiP4ndi**)But 100 * 2.079/3.02 = 64.96875 (which would round up to 65)

Thats why you are getting 64.9...% instead of 68.8%

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#15

**3.2**g, not

**3.02**g.

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

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#16

(Original post by

The mass of the tablet is

So the % mass is 2.079/

You got 68.8...% because you did 2.079/

**dip0**)The mass of the tablet is

**3.2**g, not**3.02**g.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
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#17

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#18

(Original post by

sorry im commenting late but its 2HCl so the ratio isnt 1:1 but 2mol hcL:1mol NaHCO3 so the answer is right

**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

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