In: Chemistry
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)
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%