Question

Calculate the number of moles of Ag^+ in 6.0 mL of 0.004M AgNO_3 and the number of moles of CrO_4^- in 6.0 mL of 0.0024 M K_2CrO_4.

Answer 1

mol of K2CrO4 = MV = 6*0.0024 = 0.0144 mol of K2CrO4

mol of AgNO3 = MV = 6*0.004 = 0.024 mol of AgNO3

then

only AgNO3 contains Ag+

then

0.024 *10^-3 mol of AG+ = 0.000024 mol of Ag+

ForK+

K = MV = 0.0144 *2 = 0.0288 mol of K+