Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: AshleyElizabeth on November 29, 2007, 12:08:15 AM
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This is for my CHM 130 lab, I'm having a hard time understnading, but I'm working on figuring it out through notes. There is no where in the book w/ te correct answers so I'm hoping someone can give them to me to make sure I'm doing this right. I'll check back in a few hours after I've done the problems. Thanks in advance for any help you can give. Much appericated.
Ashley
Determine the molarity of hydrocholric acid solution is exactly 34.21 mL of 0.0431 M sodium hydroxide are required to neutralize 26.34 mL of the acid solution
Determin the molarity of a Barium Hydroxide solution if 41.65 mL of it are required to neutralize 1.190 g of Potassium Hydrogen Phthalate via the following equation.
Ba(OH)2+2KHC8H404=2H20+K2C8H404
Determine the molarity of a Phosphoric Acid solution if exactly 34.21 mL of 0.0431 M Sodium Hydroxide are required to neutralize 26.34 mL of the acid solution.
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In all cases the process is the same.
Eg. In the first one, n=cv (v in L) will give the amount of NaOH reacted, then the amount of HCl nuetralised can be found from looking at rxn eqn:
NaOH + HCl ---> NaCl + H20
From rxn eqn, amount of NaOH is the same as the amount of HCl that reacts.
c=n/v can then be used to find molarity of NaOH solution since you know n and v.
There is no where in the book w/ te correct answers so I'm hoping someone can give them to me to make sure I'm doing this right.
Post your answers and then we will check them :)
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OK- n is Molarity (right?)
which would make V volume
is R the gas constant thing?
0.08206 atm x L/ moles x K?
what is rxn eqn?
Sorry, I'm really not good at this-and thank you for replying with the steps to solve the problem.
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and what is c?
sorry!
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OK
for the first problem I posted "Determine the molarity of hydrocholric acid solution is exactly 34.21 mL of 0.0431 M sodium hydroxide are required to neutralize 26.34 mL of the acid solution"
Is the correct answer 17.87 M HCl?
THis is the formula I used: M2=V2/M1V1
which would be:
M2= 26.34 mL/0.0431 M x 34.21 mL
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OK- n is Molarity (right?)
No, n is the number of moles, c is concentration in mol/L (molarity) and v is volume in L.
what is rxn eqn?
Reaction equation
THis is the formula I used: M2=V2/M1V1
Use my one ;)
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THis is the formula I used: M2=V2/M1V1
While this is technically correct, "using" "special formula" leads nowhere in this case - as it doesn't reflect stoichiometry of the reaction, thus it won't work for 2nd and 3rd question. But all these questions are identical and easy - they are just stoichiometry problems that you solve using balanced reaction equations.
Let's see how it works for the Ba(OH)2 and KHP. First - balanced reaction equation:
Ba(OH)2 + 2KHC8H4O4 -> BaC8H4O4 + 2H2O + K2C8H4O4
For us now the only important conclusion of this equation is that barium hydroxide reacts with hydrogen phtalate 1:2 - 1 mole of hydroxide with 2 moles of hydrogen phtalate. You were told there was 1.190g of potassium hydrogen phtalate - convert it to moles (it should give you 9.304x10-3 mole). As they react 1:2 hydrogen phtalate reacts with half the amount of barium hydroxide - so there was 9.304x10-3/2 = 4.653x10-3 mole of barium hydroxide. Now - you know number of moles and you know volume, as it was given to you - you have everything needed to calculate concentration, just use definition.
"Formula" you have tried to use won't work here, as it is a shortcut that can be used only for reactions where the stoichiometry is 1:1.
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M2=M1V1/V2 not V2/M1V1 :)
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Oops. So it wasn't even "technically" correct :)