Calculating Moles & Molarity of KHP & NaOH

[Prone17]

what i did:

1) i put 0.737 g of KHP in a flask
2) then added ~25 mL of distilled water
3) and then titrated 34.06 mL of NaOH into it to get its equilibrium point,

how would i find: 1. the moles of KHP, and 2. the molarity of NaOH? and would the molar mass have any significance?

1. the moles of KHP
i thought i would turn .737 g into moles by dividing by the masses given in the periodic table (K=39.1g, H=1.01g, P=30.9g) = 71.08g in total

which gives 0.0104 moles of KHP

2. the molarity of NaOH

since I am pretty sure the mole ratio of NaOH and KHP is 1:1, i used 0.0104 moles of NaOH

M=moles/L
M= 0.0104/.03406
M= .305 M

is all this correct? and if so, why did they ask me for the molar mass of KHP (which i googled to be 204.22 g/mol) ? how is this even significant?

also, when doing the molarity part, would i have to take into consideration the 25mL of distilled water i added at the beginning of the experiment, or just the amount of NaOH i added?

 

[sjb-2812]

what i did:

1) i put 0.737 g of KHP in a flask
2) then added ~25 mL of distilled water
3) and then titrated 34.06 mL of NaOH into it to get its equilibrium point,

how would i find: 1. the moles of KHP, and 2. the molarity of NaOH? and would the molar mass have any significance?

1. the moles of KHP
i thought i would turn .737 g into moles by dividing by the masses given in the periodic table (K=39.1g, H=1.01g, P=30.9g) = 71.08g in total

which gives 0.0104 moles of KHP

2. the molarity of NaOH

since I am pretty sure the mole ratio of NaOH and KHP is 1:1, i used 0.0104 moles of NaOH

M=moles/L
M= 0.0104/.03406
M= .305 M

is all this correct? and if so, why did they ask me for the molar mass of KHP (which i googled to be 204.22 g/mol) ? how is this even significant?

also, when doing the molarity part, would i have to take into consideration the 25mL of distilled water i added at the beginning of the experiment, or just the amount of NaOH i added?

I don't think the KHP you're using here has any phosphorus in it, rather it's an acid salt as you later discover

 

[Prone17]

I don't think the KHP you're using here has any phosphorus in it, rather it's an acid salt as you later discover

i'm not sure what that means. I'm just going by my lab book. i just know that the experiment was to titrate NaOH into KHP, and as far as i know, this was not a trick question. the bottle i took the KHP from said "KHP" on it.

 

[sjb-2812]

When you calculated the number of moles of KHP in

1. the moles of KHP
i thought i would turn .737 g into moles by dividing by the masses given in the periodic table (K=39.1g, H=1.01g, P=30.9g) = 71.08g in total

you made an assumption that the compound in question was literally KHP (one atom of potassium, one of hydrogen, and one of phosphorus). However, the molecular mass you got from google is not consistent with this. KHP can mean different things, and it's not totally your fault. It seems to me you are doing something similar to http://www.google.co.uk/search?q=ktp+standardization+naoh

 

[Prone17]

When you calculated the number of moles of KHP in


you made an assumption that the compound in question was literally KHP (one atom of potassium, one of hydrogen, and one of phosphorus). However, the molecular mass you got from google is not consistent with this. KHP can mean different things, and it's not totally your fault. It seems to me you are doing something similar to http://www.google.co.uk/search?q=ktp+standardization+naoh

yes i did assume that. but then i realized that to find the moles of KHP (part 1 of my question), i would divide the mass of KHP (0.737 g) by the molar mass of KHP (204.22 g/mol). is this correct? and yes, that link is exactly what i did in the experiment.

 

[Borek]

Compare:

acid base titration solution standardization

Yes, 0.737/204.22 will give you moles, yes reaction is 1:1.

 

[symbolipoint]

Just a guess: Are you abbreviating Potassium Biphthalate with "KHP"?