Answer to Question #126045 in General Chemistry for mike

Question

A 0.9030 gram sample of M(OH)2(aq) was mixed with 20.00mL of 2.000M HCl and enough water added to make 100.00mL of solution.
M(OH)2(aq) + 2HCl(aq) ⇨ MCl2(aq) + 2H2O(l)
A 10.00mL aliquot of the solution was taken and titrated with 17.645mL of 0.05121M NaOH.
Identify the metal M.

Answer 1

Milli Moles of NaOH in 17.645 mL of 0.05121 M NaOH

=

( 17.645 * 0.05121 ) = 0.9036 millimoles .

Millimole in 10 ml of HCl = 40 * 10 /100 = 4 millimole..

1 Millimole of HCl React With 1 Millimole of NaOH .

So , 0.9063 millimole of NaOH React With 0.9063 millimole of HCl .

So , Rest millimole of HCl which is React With M(OH)₂ = ( 4 - 0.9036 ) millimole . = 3.0964 millimole .

For 2 millimole of HCl 1 millimole of M( OH ) ₂ . Required .

So , 3.0964 millimole of HCl , ( 3.0964 / 2 ) millimole of M( OH )₂ Required .

So , moles of M( OH )₂ Reacted =( 3.0964 * 10^-3 ) / 2 . mole . .......................1.

Let Atomic Mass of Metal is " A ".

MOLAR MASS OF M ( OH)₂ = A + 34 ..

Moles of M ( OH )₂ = 0.9030 gram / ( A+ 34 ) gram / mole . .......................2.

From Equation 1 and 2 .

Moles of M( OH) ₂ = ( 3.0964 * 10^-3 ) / 2 = 0.9030 gram / ( A+ 34 ) gram / mole .

SO , A = 56

Metal is Fe .