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.
[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)
work:
moles of ammonium in sample:
(6.30x10^-3)(.1262)
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!
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