Question

# [Ni(NH3)m(H2O)p]^2+ + mHCl ->mCl- + pH2O + Ni^2+ + mNH4^+ excess HCl + OH- -> H2O...

[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) B. Propose a molecular formula for the salt that is consistent with this experimental equivalent weight.

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 0.1262 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.

the weight of the salt is 183.23 g

At any rate, the following is a proposal which is consistent with the data given.

Ni (58.69) + 1 NH3 (17.03) + 5 H2O (4 X 18.015 = 72.06) + 1 Cl- (35.45) and you get 183.23 g . m= 1 and p =4
The formula consistent with these data is [Ni(NH3)1(H2O)5]Cl^+1.

remaining 1.77 g due to measurement errors only.

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