Answer to Question #61012 in Chemistry for Tebogo

Question

A 0,2538 g sample contains a mixture of only KBr and KI.The sample is treated with AgNO3 and the precipitate of AgBr and AgI dried and weighed.If the precipitate mass is 0,3801 g ,what is the percentage of KBr in the original sample?

Answer 1

Let’s assume that the percentage of KBr is Z %. Then masses and amounts of KBr and KI equal:
m(KBr) = (0.2538×z)/100 = 2.538×10-3z
ν(KBr) = m(KBr)/M(KBr) = (2.538×10-3z)/119
m(KI) = 0.2538 - 2.538×10-3z
ν(KI) = m(KI)/M(KI) = (0.2538 - 2.538×10-3z)/166
Each salt forms precipitate with AgNO3 by the reaction:
KE + AgNO3 → AgE + KNO3 , where E = Br or I
Amounts and masses of AgI and AgBr can be found as follows:
ν(AgBr) = ν(KBr), thus
m(AgBr) = ν(AgBr)× M(AgBr) = [(2.538×10-3z)/119] × 188 = 4.01×10-3z
ν(AgI) = ν(KI), thus
m(AgI) = ν(AgI)× M(AgI) = [(0.2538 - 2.538×10-3z)/166] × 235 = 0.3593 – 3.59×10-3z
Taking into account that
m(AgBr) + m(AgI) = 0.3801
Solving the following equation gives the value of Z:
4.01×10-3z + 0.3593 – 3.59×10-3z = 0.3801
0.42×10-3z = 0.0208
z = 49.52 %