Chemistry question help! READ THIS (1 Viewer)

[daisy_22]

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]

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:



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]

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

 

[daisy_22]

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]

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