Answer to Question #99697 in Physical Chemistry for vats

Question

50 ml of a 0.1m koh solution is added to 25ml of 0.2m hcl. the volume of the resulting solution was made upto 500 ml. now 0.745 gram of solid kcl is added to solution . what is the concentration of cl- in the final solution . what volume of 0.1 m agno3 would be requried for complete precipitaion of cl - ions as silver chloride

Answer 1

Solution.

$KOH + HCl = KCl + H2O$

$n(KOH) = C(KOH) \times V(KOH)$

$n(HCl) = C(HCl) \times V(HCl)$

n(KOH) = 0.005 mol

n(HCl) = 0.005 mol

n(KOH) = n(HCl) = n(KCl) = 0.005 mol

$m(KCl) = 0.005 \times 74.5513 = 0.373 \ g$

$m'(KCl) = 0.373 + 0.745 = 1.118 \ g$

$KCl = K^+ + Cl^-$

C(KCl) = C(Cl-)

$C'(KCl) = \frac{m'(KCl)}{M(KCl) \times V}$

$C'(KCl) =\frac{1.118}{74.5513 \times 0.5} = 0.03 M$

C(Cl-) = 0.03 M

$AgNO3 + KCl = AgCl + KNO3$

$C(AgNO3) \times V(AgNO3) = C(KCl) \times V(KCl)$

$V(AgNO3) = \frac{C(KCl) \times V(KCl)}{C(AgNO3)}$

$V(AgNO3) = \frac{0.03 \times 0.5}{0.1} = 0.15 L = 150 \ ml$

Answer:

C(Cl-) = 0.03 M

V(AgNO3) = 150 ml = 0.15 L