A 0.2368 sample contained only NaCl and KBr. It was dissolved in water and required 44.80...

A 0.2368 sample contained only NaCl and KBr. It was dissolved in water and required 44.80 mL of 0.04873 M AgNO3 for complete titration of both halides. Calculate the weight percent of Br in the solid sample. (answer is 60.85 but please show work)

Solutions

Expert Solution

Molar mass of NaCl = 58.44g/mol

Molar mass of KBr = 119.00 g/mol

Let's say mass of NaCl = x

Mass of KBr =0.2368-x

Moles of NaCl = x/58.44

Moles of KBr = (0.2368-x)/119.00

KBr + AgNO3 = AgBr + KNO3

NaCl + AgNO3 = NaNO3 + AgCl

AgNO3 is completely consumed by these two reactions combined

Moles of AgNO3 used = molarity*volume in litres

= 0.04873M*0.0448L

= 0.002183 moles

Which is equal to moles of KBr+NaCl as seen from reaction equation that each of them need equavalent amount of AgNO3

0.002183 moles =x/58.44 + (0.2368-x)/119.00

15.18 = 119x +13.84 -58.44x

1.34 = 60.56x

x = 0.02213 g

Mass of KBr =0.2368-0.02213

=0.21467g

119.00 g KBr has 79.90g Br

0.21467g KBr has 0.14414 g Br

%Br = mass of Br/total mass *100

= 0.14414/0.2368 *100 = 60.86%

queen_honey_blossom answered 1 year ago

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