[jpuppy21]

Hey I've been trying for hours to find out how to figure this question out:

A 0.6142 gram sample of Ni(OH)X and 20.00 mL of 2.000 M HCl were put into a 100.0 mL volumetric flask and mixed with enough extra water to make 100.0 mL of solution. A 10.00 mL aliquot of this solution was taken and titrated with 17.64 mL of 0.1516 M NaOH. What is the empirical formula of the nickel hydroxide?
HCl(aq) + OH-(aq) Cl-(aq) + H2O(l)
[Ans.: Ni(OH)2]

Any help would be appreciated thanks!

[Borek]

Start with both reaction equations.

First reaction uses excess HCl (Ni(OH)_x is a limiting reagent).

Second reaction neutralizes excess HCl.

Calculate excess HCl, then calculate amount of HCl that reacted with Ni(OH)_x.

[jpuppy21]

Thanks for the help but I still cant come up with the right answer...

1.    .02L HCl x 2.0M = .04mol HCl reacted
2.   .01764L NaOH x .1516M = .00267 mol HCl excess
3.   .04mol - .00267mol = .0133 mol HCl
4.   .0133 mol HCl x 1mol OH-/1mol HCl = .0133 mol OH-
5.   diluted:   .0133 mol in 10 ml -->  .133 mol in 100 ml
6.   0.133 mol OH- x .6142g/1mol = .0817 g OH
7.   .6142g total - .0817g OH = .53 g Ni
8.   0.53 g Ni x 1mol/59g = .009 mol Ni
9.   (.009 mol Ni :   0.133 mol OH) / .009 =
      1 mol Ni : 15 mol
?

[Borek]

Looks to me like you have accounted for dilution in a wrong moment.

[DrCMS]

Thanks for the help but I still cant come up with the right answer...
1.    .02L HCl x 2.0M = .04mol HCl reacted

Correct

2.   .01764L NaOH x .1516M = .00267 mol HCl excess
3.   .04mol - .00267mol = .0133 mol HCl
4.   .0133 mol HCl x 1mol OH-/1mol HCl = .0133 mol OH-
5.   diluted:   .0133 mol in 10 ml -->  .133 mol in 100 ml

No 0.1764*0.1516 does equal 0.00267 but that is from 10ml so in 100ml there was 0.0267mol HCl.  
That means in the original solution that there was 0.04-0.0267 = 0.0133mol OH.

6.   0.133 mol OH- x .6142g/1mol = .0817 g OH
7.   .6142g total - .0817g OH = .53 g Ni
8.   0.53 g Ni x 1mol/59g = .009 mol Ni
9.   (.009 mol Ni :   0.133 mol OH) / .009 =
      1 mol Ni : 15 mol

No your maths is wrong 0.0133mol OH = 0.2261g OH (0.0133*17)
0.6142g total - 0.2261g OH = 0.3881g Ni
0.3881g Ni = 0.0066mol Ni
0.0066 mol Ni : 0.0133 mol OH = NiOH₂