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Chemistry question help! READ THIS (1 Viewer)

daisy_22

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The titration between potassium hydrogen phthalate and sodium hydroxide is a 1:1 stoichiometry. If the mean titration volume (titre) of sodium hydroxide to reach the endpoint is 26.2 mL against 25 mL of a 7.06 mM standard solution of potassium hydrogen phthalate, what is the concentration of sodium hydroxide (in mM)? Provide your answer to 3 significant figures.

Note: this is a hypothetical scenario and the concentrations may not match the previous question or the values used in your experiment.

I kept on doing this, and got the answer to be 7.404nM, but the quiz says its wrong, can someone provide me with the right answer and explanation? Thanks in advance
 

jimmysmith560

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There exists a very similar version of this question, where the only difference is in some of the values ("22.7 mL against 25 mL of a 7.33 mM standard solution...").

The following working is targeted at the version of this question with the different values:

1684676218215.png

Perhaps it would still be useful in assisting you to solve your own version of the question, since the only difference is in the given values.

I hope this helps! 😄
 

carrotsss

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Moles of potassium hydrogen phthalate = c * v
= 0.025*7.06
=0.1765 mol
Therefore there are also 0.1765 mol of sodium hydroxide (1:1 ratio)
concentration=n/v
=0.1765/0.0262
=6.74 M (3sf)

It seems like you’ve got confused, in the question the 7.06M and 25mL is of potassium hydrogen phthalate, not sodium hydroxide like usual
 
Last edited:

daisy_22

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There exists a very similar version of this question, where the only difference is in some of the values ("22.7 mL against 25 mL of a 7.33 mM standard solution...").

The following working is targeted at the version of this question with the different values:

View attachment 38463

Perhaps it would still be useful in assisting you to solve your own version of the question, since the only difference is in the given values.

I hope this helps! 😄
Thank you so much!
 

daisy_22

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Moles of potassium hydrogen phthalate = c * v
= 0.025*7.06
=0.1765 mol
Therefore there are also 0.1765 mol of sodium hydroxide (1:1 ratio)
concentration=n/v
=0.1765/0.0262
=6.74 M (3sf)

It seems like you’ve got confused, in the question the 7.06M and 25mL is of potassium hydrogen phthalate, not sodium hydroxide like usual
Yeah I did :(
 

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