Please help me!! I can’t find the answer. Please help me!! I can’t find the answ

Question

Please help me!! I can’t find the answer. Please help me!! I can’t find the answer. Please help me!! I can’t find the answer. 13-17. An analysis for borohydride ion is based upon its reaction with Ag: BH: 8Agt +80H-H2BO5 8Ag(s) 5H,0 The purity of a quantity of KBH4 for use in an organic synthesis was established by diluting 3.213 g of the mate- rial to exactly 500.0 mL, treating a 100.0-mL aliquot with 50.00 mL of 0.2221 M AgNO,, and titrating the excess silver ion with 3.36 mL of 0.0397 M KSCN. Calculate the percent purity of the KBH (53.941 g/mol).

Explanation / Answer

Millimoles of Ag+ added (as AgNO3) = (50.00 mL)*(0.2221 M) = 11.105 mmole.

Ag+ reacts with SCN- (added as KSCN) as below.

Ag+ (aq) + SCN- (aq) -----------> AgSCN (s)

As per the stoichiometric equation,

1 mole Ag+ = 1 mole SCN-.

Millimoles of SCN- added = millimoles of Ag+ reacted with SCN- = (3.36 mL)*(0.0397 M) = 0.133392 mmole.

Millimoles of Ag+ reacted with BH4- added as KBH4 = (11.105 – 0.133392) mmole = 10.971608 mmole.

As per the provided stoichiometric equation in the question,

1 mole BH4- = 8 moles Ag+.

Therefore, 10.971608 mmole Ag+ = (10.971608 mmole Ag+)*(1 mole BH4-/8 moles Ag+) = 1.371451 mmole BH4-.

Now, 100.00 mL aliquot was taken from the prepared 500.00 mL sample solution; hence, the millimoles of BH4- present in 500.00 mL sample solution = (1.371451 mmole)*(500.00 mL)/(100.00 mL) = 6.857255 mmole.

Mass of KBH4 corresponding to 6.857255 mmole = (6.857255 mmole)*(1 mole/1000 mmole)*(53.941 g/1 mole) = 0.369887 g.

Percent purity of KBH4 in the sample = (0.369887 g)/(3.213 g)*100 = 11.5122% 11.51% (ans).

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