A mineral contains 11.11% Ni (58.71 g/mol); 0.300 g of the same is treated and its...

A mineral contains 11.11% Ni (58.71 g/mol); 0.300 g of the same is treated and its ammoniacal solution is treated with 50.00 mL of a solution of 0.0911 M KCN. A little KI is added as an indicator and the solution is titrated with 0.0903 M AgNO₃ until slight turbidity appears. What volume of AgNO₃ will be needed?

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Expert Solution

Ni mass in sample = ( mass of sample ) x ( percentage of Ni / 100)

= ( 0.3g x 11.11 /100) = 0.0333 g

Ni moles = mass / atomic mass of Ni = 0.0333 g / 58.71 g/mol = 0.0005677 moles

KCN moles = M x V ( inL) = 0.0911 x ( 50/1000) = 0.004555 = CN- moles ( since 1KCN gives 1CN-)

CN- combines with Ni in 4:1 ratio to form [Ni(CN)4]^2- complex

hence CN- moles consumed = 4 x Ni moles = 4 x 0.0005677 = 0.0022708

CN- moles left unreacted = initial CN- moles - CN moles consumed

= 0.004555 - 0.0022708 = 0.002284

Ag+ reacts with CN- in 1:1 to form pecipitate AgCN

hence Ag+ moles needed to react with CN- moles = 0.002284

Ag+ is giveb by AgNO3 , hence AGNO3 moles = 0.002284

Molarity of AgNO3 = moles / volume

0.0903 = 0.002284 / volume

volume = 0.0253 L = 25.3 ml is volume of AgNO3 needed

queen_honey_blossom answered 1 year ago

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