Answer
a. $ 90.0 \space mL \space NaOH \space solution$
b. $ 2.84 \space g \space Ni(OH)_2 $
c. $0.0417 \space M \space NiCl_2$
Work Step by Step
a.
1. Find the conversion factors
- 1 L = 1000 mL
- 0.500 M $NiCl_2$ solution: $$0.500 \space mole \space NiCl_2 = 1 \space L \space solution$$
- $ 2 \space moles \space NaOH = 1 \space mole \space NiCl_2 $
- 0.200 M $NaOH$ solution: $$0.200 \space mole \space NaOH = 1 \space L \space solution$$
2. Calculate the volume of the other solution:
$$ 18.0 \space mL \space NiCl_2 \space solution \times\frac{1 \space L}{1000 \space mL} \times\frac{0.500 \space mole \space NiCl_2 }{1 \space L \space NiCl_2 \space solution} \times \frac{ 2 \space moles \space NaOH }{ 1 \space mole \space NiCl_2 } \times \frac{1 \space L \space NaOH \space solution}{0.200 \space mole \space NaOH}\times \frac{1000 \space mL}{1 \space L} = 90.0 \space mL \space NaOH \space solution$$
b.
1. Find the conversion factors
- 1 L = 1000 mL
- 1.75 M $NaOH$ solution: $$1.75 \space moles \space NaOH = 1 \space L \space solution$$
- $ 1 \space mole \space Ni(OH)_2 = 2 \space moles \space NaOH $
$ Ni(OH)_2 $ : ( 58.69 $\times$ 1 )+ ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 92.71 g/mol
$$ \frac{1 \space mole \space Ni(OH)_2 }{ 92.71 \space g \space Ni(OH)_2 } \space and \space \frac{ 92.71 \space g \space Ni(OH)_2 }{1 \space mole \space Ni(OH)_2 }$$
2. Calculate the mass in g of $Ni(OH)_2$:
$$ 35.0 \space mL \space NaOH \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{1.75 \space moles \space NaOH }{1 \space L \space NaOH \space solution} \times \frac{ 1 \space mole \space Ni(OH)_2 }{ 2 \space moles \space NaOH } \times \frac{ 92.71 \space g \space Ni(OH)_2 }{1 \space mole \space Ni(OH)_2 } = 2.84 \space g \space Ni(OH)_2 $$
c.
1. Find the conversion factors
- 1 L = 1000 mL
- 0.250 M $NaOH$ solution: $$0.250 \space mole \space NaOH = 1 \space L \space solution$$
- $ 1 \space mole \space NiCl_2 = 2 \space moles \space NaOH $
$ NiCl_2 $ : ( 35.45 $\times$ 2 )+ ( 58.69 $\times$ 1 )= 129.59 g/mol
$$ \frac{1 \space mole \space NiCl_2 }{ 129.59 \space g \space NiCl_2 } \space and \space \frac{ 129.59 \space g \space NiCl_2 }{1 \space mole \space NiCl_2 }$$
2. Calculate the amount in moles of $NiCl_2$:
$$ 10.0 \space mL \space NaOH \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{0.250 \space mole \space NaOH }{1 \space L \space NaOH \space solution} \times \frac{ 1 \space mole \space NiCl_2 }{ 2 \space moles \space NaOH } = 0.00125 \space mole \space NiCl_2 $$
3. Calculate the molarity:
$$Volume(L) = 30.0 \space mL \times \frac{1 \space L}{1000 \space mL} = 0.0300 \space L $$
$$Molarity(M) = \frac{0.00125 \space mole\space NiCl_2}{0.0300 \space L} = 0.0417 \space M \space NiCl_2$$
