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miNi
Korea (South)2010 Posts
1. calculate the molarity of a solution of sodium hydroxide if 23.64mL of this solution is needed to neutralize .5632g of potassium hydrogen phthalate. 2. It is found that 24.68mL of .1165M NaOH is needed to titrate .2931 g of an unknown acid to the phenolphthalein end point. Calculate the equivalent mass of the acid. help appreciated  | ||
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GrayArea
United States872 Posts
Second one I am not really sure.  | ||
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miNi
Korea (South)2010 Posts
it's an acid-base neutralization exercise | ||
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GrayArea
United States872 Posts
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miNi
Korea (South)2010 Posts
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SIUnit
China288 Posts
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SayaSP
Laos5494 Posts
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miNi
Korea (South)2010 Posts
cant find out the [H+] because KHP is a weak acid | ||
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SIUnit
China288 Posts
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SayaSP
Laos5494 Posts
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GrayArea
United States872 Posts
in my defense, i took general chem 2 years ago, so ya..but still, damn...sry man | ||
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SIUnit
China288 Posts
Ok, for the first question, ignore my first response. At the endpoint, the number of moles of NaOH used in the titration is equal to the number of moles of KHP (Molar mass 204.2 g/mole) weighed out. Knowing the volume of base and number of moles of acid, you can calculate the concentration of the NaOH solution using the following expression Macid * Vacid = Mbase * Vbase Macid = molarity (moles/L) of acid Vacid = volume (L) of acid Mbase = molarity (moles/L) of base Vbase = volume (L) of base Which is basically the moles of KHP over liters of NaOH used. While KHP is a weak acid, it does not change the fact that when the end point is reached, theoretically the moles of OH- should equal the moles of H+ released from the acid. | ||
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SayaSP
Laos5494 Posts
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Steelflight-Rx
United States1389 Posts
On February 25 2008 14:32 GrayArea wrote: dude, im taking organic chemistry in college, i should know this stuff but i dont. -_- rofl same here, i was just thinking that to myself | ||
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xhuwin
United States476 Posts
Well I'm gonna assume the acid is monoprotic. Anyway this is just stoichiometry: Moles Base = Moles Acid at the titration end point. Moles Base = .02468 L * .1165 M NaOH = .2931 g Acid / (X g per mol Acid) ---- x is what you're solving for, the molar mass of the acid. Basically, for the base, we multiply volume by concentration, and for acid, we divide mass by molar mass. Both sides of the equation give us a number of moles, which we know is equal because we're talking about the titration end point (assuming a monoprotic acid). PM me if you still need help! | ||
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fight_or_flight
United States3988 Posts
On February 25 2008 14:37 SIUnit wrote: Macid * Vacid = Mbase * Vbase This should be the only equation you need I'm pretty sure. | ||
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