Question

A 750.25 gram alloyof nickel was dissolved and treated to remove impurities. Its ammoniacal solution was treated with 50 ml of 0.1075 M KCN and the excess cyanide required 2.25 ml of 0.00925 M AgNO₃. Determine the % Ni in the alloy ? Answer: 10.53 %

Answer 1

50 ml = 0.050 L

KCN < => K+   + CN-

0.050 L of 0.1075 mol/Litre KCN = 0.005375 moles of CN-

2.25 ml = 0.00225 L

0.00225 L of 0.00925 mol/L AgNO3= 0.000021 AgNO3 moles

(0.005375 mol of CN- ) – (0.000021 AgNO3 mol )

= 0.005354 mol CN

1 mol Ni2+   +   4 CN-    --> one mole [Ni(CN)4] 2-

0.005354 moles of CN- * 1 Ni /4 CN-

= 0.002677 moles of Ni
now calcualte the mass of 0.0013385 moles of Ni :


0.0013385 moles of Ni * 58.69 g/mol = 0.0785556565 g Ni

750.25 milligram = 0.75025 g

0.0785556565 g Ni / 0.75025 g sample *100= 10.47 or 10.5 %