Question

Barium iodate is a white, granular inorganic compound that is stable on a temperature that is approximately 580 °C (1,076 °F). It is used as an additive to salt and bread without notable toxic effects. If a 50.00 mL aliquot of a solution containing 0.1998 g of BaCl2•2H2O is mixed with 50.00 mL aliquot of a solution containing 0.3005 g NaIO3. Assume that the solubility of the Ba(IO3)2 in water is negligible,

a. How many barium and iodate ions are contained in their original solutions? (5 points)

b. What is the mass of the precipitate formed? (5 points)

c. Calculate for the percentage of the unreacted compound that remained in the

solution.

Answer 1

$M(BaCl_2*2H_2O)=137+35.5*2+2*18=244g/mol$

$M(NaIO_3)=23+127+3*16=198g/mol$

$M(Ba(IO_3)_2)=137+(127+3*16)*2=487g/mol$

a. $\nu(BaCl_2*2H_2O)=\nu(Ba2+)=0.1998/244 mol=0.00082 mol$

$\nu(NaIO_3)=\nu(IO_3^-)=0.3005/198 mol=0.00152 mol$

b.$Ba(2+)+2IO_3^-=Ba(IO_3)_2$

iodate ions deficient, so $\nu(Ba(IO_3)_2)=\nu(NaIO_3)/2=0.00076mol$

$m(Ba(IO_3)_2)=\nu(Ba(IO_3)_2)*M(Ba(IO_3)_2)=0.37g$

c. Unreacted $BaCl_2*2H_2O$ :

$\Delta\nu(BaCl_2*2H_2O)=\nu(BaCl_2*2H_2O)-\nu(NaIO_3)/2=6*10^(-5)mol$

$\Delta m(BaCl_2*2H_2O)=M(BaCl_2*2H_2O)*\nu(BaCl_2*2H_2O)=0.01464g$

$W(BaCl_2*2H_2O)=\Delta m(BaCl_2*2H_2O)/(M(BaCl_2*2H_2O)*\nu(BaCl_2*2H_2O))=$

$0.01464/(244*0.00082)=0.073$

7.3 %