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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.
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    (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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    (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.
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    (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%
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    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.
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    (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
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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 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.
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    (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.
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    (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%
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    (Original post by charco)
    actually I meant 68.8%
    But 100 * 2.079/3.2 = 64.96875 (which would round up to 65)
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    (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 ...

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    (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%
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    (Original post by charco)
    Not on my calculator it's not ...

    :dontknow:
    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
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    (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 ...
 
 
 
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