Posted by Anonymous on Sunday, October 4, 2009 at 3:07am.
[Ni(NH3)m(H2O)p]^2+ + mHCl ->mCl- + pH2O + Ni^2+ + mNH4^+
excess HCl + OH- -> H2O
A .185 g sample of nickel salt was dissolved in 30.00 mL of 0.1013 N HCl. The excess HCl required 6.30 mL of 0.1262 N NaOH to reach the end point. Calculate the weight of the salt that contains one mole of NH3 (that is the equivalent weight of the salt)
moles of ammonium in sample:
weight of salt that contains one mole of NH3: 0.185/(6.30x10^-3)(.1262)
Propose a molecular formula for the salt in 3 that is consistent with this experimental equivalent weight.
I don't know if I did the first problem correctly. I'm not sure how to do the second problem (proposing the molecular formula - the hint the teacher gave us is that the sum of m and p is less than or equal to 6 and also bromine atoms will need to be mentioned). Thanks in advance for the help!
chemistry (drbob222) - DrBob222, Sunday, October 4, 2009 at 12:49pm
I don't think you did the first part correctly.
# milliequivalents = me = mL x N
milliquivalent weight = mew
me HCl initially added = 30 mL x 0.1013 N = 3.039 me.
me NaOH used to titrate the excess HCl added = 6.30 mL x 0.1015 N = 0.7951 me.
me HCl used in the reaction = 3.039 - 0.7951 = 2.244 me.
Then 0.185/2.244 = 0.08244 = mew of the salt = 0.08244 or an equivalent weight of 82.44.
There is nothing in the problem, that I see, to indicate that the number of NH3 molecules is 2; since you refer to problem 3 I suspect you may have omitted something. At any rate, the following is a proposal which is consistent with the data given.
Since the equivalent weight is 82.44, the molar mass is twice that if the ammonia is 2 molecules to the formula weight, so 82.44 x 2 = 164.88.
Subtract Ni at 58.69 and 2NH3 (2 x 17.03) and that leaves 72.13. Divide that by the molar mass of H2O (18.015) and you get 4.0 so that is p.
The formula consistent with these data is [Ni(NH3)2(H2O)4]^+2. Check my thinking.
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